Solve (x-1/2)^2-(2x-3/2-1)^2/2-frac{2/3-1}{frac{1}{6}-frac{1}{2}}=2(x-2)

$$\frac { ( x - \frac { 1 } { 2 } ) ^ { 2 } - ( \frac { 2 x - 3 } { 2 } - 1 ) ^ { 2 } } { 2 } - \frac { \frac { 2 } { 3 } - 1 } { \frac { 1 } { 6 } - \frac { 1 } { 2 } } = 2 ( x - 2 )$$

Solve for x (complex solution)

$$x\in \mathrm{C}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$\left(x-\frac{1}{2}\right)^{2}-\left(\frac{2x-3}{2}-1\right)^{2}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{1}{2}\right)^{2}.

$$x^{2}-x+\frac{1}{4}-\left(\frac{2x-3}{2}-1\right)^{2}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{2x-3}{2}-1\right)^{2}.

$$x^{2}-x+\frac{1}{4}-\left(\left(\frac{2x-3}{2}\right)^{2}-2\times \frac{2x-3}{2}+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

To raise \frac{2x-3}{2} to a power, raise both numerator and denominator to the power and then divide.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}-2\times \frac{2x-3}{2}+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Express -2\times \frac{2x-3}{2} as a single fraction.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}+\frac{-2\left(2x-3\right)}{2}+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Cancel out 2 and 2.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}-\left(2x-3\right)+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

To find the opposite of 2x-3, find the opposite of each term.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}-2x+3+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Add 3 and 1 to get 4.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}-2x+4\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -2x+4 times \frac{2^{2}}{2^{2}}.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}+\frac{\left(-2x+4\right)\times 2^{2}}{2^{2}}\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Since \frac{\left(2x-3\right)^{2}}{2^{2}} and \frac{\left(-2x+4\right)\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$x^{2}-x+\frac{1}{4}-\frac{\left(2x-3\right)^{2}+\left(-2x+4\right)\times 2^{2}}{2^{2}}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Do the multiplications in \left(2x-3\right)^{2}+\left(-2x+4\right)\times 2^{2}.

$$x^{2}-x+\frac{1}{4}-\frac{4x^{2}-12x+9-8x+16}{2^{2}}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Combine like terms in 4x^{2}-12x+9-8x+16.

$$x^{2}-x+\frac{1}{4}-\frac{4x^{2}-20x+25}{2^{2}}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}-x+\frac{1}{4}-\frac{4x^{2}-20x+25}{4}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Since \frac{1}{4} and \frac{4x^{2}-20x+25}{4} have the same denominator, add them by adding their numerators.

$$x^{2}-x+\frac{1+4x^{2}-20x+25}{4}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Combine like terms in 1+4x^{2}-20x+25.

$$x^{2}-x+\frac{26+4x^{2}-20x}{4}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Subtract 1 from \frac{2}{3} to get -\frac{1}{3}.

$$x^{2}-x+\frac{26+4x^{2}-20x}{4}+6\left(-\frac{1}{3}\right)=4\left(x-2\right)$$

Multiply 6 and -\frac{1}{3} to get -2.

$$x^{2}-x+\frac{26+4x^{2}-20x}{4}-2=4\left(x-2\right)$$

Use the distributive property to multiply 4 by x-2.

$$x^{2}-x+\frac{26+4x^{2}-20x}{4}-2=4x-8$$

Divide each term of 26+4x^{2}-20x by 4 to get \frac{13}{2}+x^{2}-5x.

$$x^{2}-x+\frac{13}{2}+x^{2}-5x-2=4x-8$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-x+\frac{13}{2}-5x-2=4x-8$$

Combine -x and -5x to get -6x.

$$2x^{2}-6x+\frac{13}{2}-2=4x-8$$

Subtract 2 from \frac{13}{2} to get \frac{9}{2}.

$$2x^{2}-6x+\frac{9}{2}=4x-8$$

Subtract 4x from both sides.

$$2x^{2}-6x+\frac{9}{2}-4x=-8$$

Combine -6x and -4x to get -10x.

$$2x^{2}-10x+\frac{9}{2}=-8$$

Add 8 to both sides.

$$2x^{2}-10x+\frac{9}{2}+8=0$$

Add \frac{9}{2} and 8 to get \frac{25}{2}.

$$2x^{2}-10x+\frac{25}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -10 for b, and \frac{25}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 2\times \frac{25}{2}}}{2\times 2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 2\times \frac{25}{2}}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-10\right)±\sqrt{100-8\times \frac{25}{2}}}{2\times 2}$$

Multiply -8 times \frac{25}{2}.

$$x=\frac{-\left(-10\right)±\sqrt{100-100}}{2\times 2}$$

Add 100 to -100.

$$x=\frac{-\left(-10\right)±\sqrt{0}}{2\times 2}$$

Take the square root of 0.

$$x=-\frac{-10}{2\times 2}$$

The opposite of -10 is 10.

$$x=\frac{10}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{10}{4}$$

Reduce the fraction \frac{10}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$\left(x-\frac{1}{2}\right)^{2}-\left(\frac{2x-3}{2}-1\right)^{2}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{1}{2}\right)^{2}.

$$x^{2}-x+\frac{1}{4}-\left(\frac{2x-3}{2}-1\right)^{2}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{2x-3}{2}-1\right)^{2}.

$$x^{2}-x+\frac{1}{4}-\left(\left(\frac{2x-3}{2}\right)^{2}-2\times \frac{2x-3}{2}+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

To raise \frac{2x-3}{2} to a power, raise both numerator and denominator to the power and then divide.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}-2\times \frac{2x-3}{2}+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Express -2\times \frac{2x-3}{2} as a single fraction.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}+\frac{-2\left(2x-3\right)}{2}+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Cancel out 2 and 2.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}-\left(2x-3\right)+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

To find the opposite of 2x-3, find the opposite of each term.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}-2x+3+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Add 3 and 1 to get 4.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}-2x+4\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -2x+4 times \frac{2^{2}}{2^{2}}.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}+\frac{\left(-2x+4\right)\times 2^{2}}{2^{2}}\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Since \frac{\left(2x-3\right)^{2}}{2^{2}} and \frac{\left(-2x+4\right)\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$x^{2}-x+\frac{1}{4}-\frac{\left(2x-3\right)^{2}+\left(-2x+4\right)\times 2^{2}}{2^{2}}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Do the multiplications in \left(2x-3\right)^{2}+\left(-2x+4\right)\times 2^{2}.

$$x^{2}-x+\frac{1}{4}-\frac{4x^{2}-12x+9-8x+16}{2^{2}}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Combine like terms in 4x^{2}-12x+9-8x+16.

$$x^{2}-x+\frac{1}{4}-\frac{4x^{2}-20x+25}{2^{2}}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}-x+\frac{1}{4}-\frac{4x^{2}-20x+25}{4}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Since \frac{1}{4} and \frac{4x^{2}-20x+25}{4} have the same denominator, add them by adding their numerators.

$$x^{2}-x+\frac{1+4x^{2}-20x+25}{4}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Combine like terms in 1+4x^{2}-20x+25.

$$x^{2}-x+\frac{26+4x^{2}-20x}{4}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Subtract 1 from \frac{2}{3} to get -\frac{1}{3}.

$$x^{2}-x+\frac{26+4x^{2}-20x}{4}+6\left(-\frac{1}{3}\right)=4\left(x-2\right)$$

Multiply 6 and -\frac{1}{3} to get -2.

$$x^{2}-x+\frac{26+4x^{2}-20x}{4}-2=4\left(x-2\right)$$

Use the distributive property to multiply 4 by x-2.

$$x^{2}-x+\frac{26+4x^{2}-20x}{4}-2=4x-8$$

Divide each term of 26+4x^{2}-20x by 4 to get \frac{13}{2}+x^{2}-5x.

$$x^{2}-x+\frac{13}{2}+x^{2}-5x-2=4x-8$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-x+\frac{13}{2}-5x-2=4x-8$$

Combine -x and -5x to get -6x.

$$2x^{2}-6x+\frac{13}{2}-2=4x-8$$

Subtract 2 from \frac{13}{2} to get \frac{9}{2}.

$$2x^{2}-6x+\frac{9}{2}=4x-8$$

Subtract 4x from both sides.

$$2x^{2}-6x+\frac{9}{2}-4x=-8$$

Combine -6x and -4x to get -10x.

$$2x^{2}-10x+\frac{9}{2}=-8$$

Subtract \frac{9}{2} from both sides.

$$2x^{2}-10x=-8-\frac{9}{2}$$

Subtract \frac{9}{2} from -8 to get -\frac{25}{2}.

$$2x^{2}-10x=-\frac{25}{2}$$

Divide both sides by 2.

$$\frac{2x^{2}-10x}{2}=-\frac{\frac{25}{2}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{10}{2}\right)x=-\frac{\frac{25}{2}}{2}$$

Divide -10 by 2.

$$x^{2}-5x=-\frac{\frac{25}{2}}{2}$$

Divide -\frac{25}{2} by 2.

$$x^{2}-5x=-\frac{25}{4}$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-\frac{25}{4}+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=\frac{-25+25}{4}$$

Add -\frac{25}{4} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-5x+\frac{25}{4}=0$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{5}{2}=0 x-\frac{5}{2}=0$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{5}{2} x=\frac{5}{2}$$

The equation is now solved. Solutions are the same.

$$x=\frac{5}{2}$$

Solve for x

$$x\in \mathrm{R}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$\left(x-\frac{1}{2}\right)^{2}-\left(\frac{2x-3}{2}-1\right)^{2}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{1}{2}\right)^{2}.

$$x^{2}-x+\frac{1}{4}-\left(\frac{2x-3}{2}-1\right)^{2}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{2x-3}{2}-1\right)^{2}.

$$x^{2}-x+\frac{1}{4}-\left(\left(\frac{2x-3}{2}\right)^{2}-2\times \frac{2x-3}{2}+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

To raise \frac{2x-3}{2} to a power, raise both numerator and denominator to the power and then divide.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}-2\times \frac{2x-3}{2}+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Express -2\times \frac{2x-3}{2} as a single fraction.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}+\frac{-2\left(2x-3\right)}{2}+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Cancel out 2 and 2.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}-\left(2x-3\right)+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

To find the opposite of 2x-3, find the opposite of each term.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}-2x+3+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Add 3 and 1 to get 4.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}-2x+4\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -2x+4 times \frac{2^{2}}{2^{2}}.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}+\frac{\left(-2x+4\right)\times 2^{2}}{2^{2}}\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Since \frac{\left(2x-3\right)^{2}}{2^{2}} and \frac{\left(-2x+4\right)\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$x^{2}-x+\frac{1}{4}-\frac{\left(2x-3\right)^{2}+\left(-2x+4\right)\times 2^{2}}{2^{2}}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Do the multiplications in \left(2x-3\right)^{2}+\left(-2x+4\right)\times 2^{2}.

$$x^{2}-x+\frac{1}{4}-\frac{4x^{2}-12x+9-8x+16}{2^{2}}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Combine like terms in 4x^{2}-12x+9-8x+16.

$$x^{2}-x+\frac{1}{4}-\frac{4x^{2}-20x+25}{2^{2}}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}-x+\frac{1}{4}-\frac{4x^{2}-20x+25}{4}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Since \frac{1}{4} and \frac{4x^{2}-20x+25}{4} have the same denominator, add them by adding their numerators.

$$x^{2}-x+\frac{1+4x^{2}-20x+25}{4}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Combine like terms in 1+4x^{2}-20x+25.

$$x^{2}-x+\frac{26+4x^{2}-20x}{4}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Subtract 1 from \frac{2}{3} to get -\frac{1}{3}.

$$x^{2}-x+\frac{26+4x^{2}-20x}{4}+6\left(-\frac{1}{3}\right)=4\left(x-2\right)$$

Multiply 6 and -\frac{1}{3} to get -2.

$$x^{2}-x+\frac{26+4x^{2}-20x}{4}-2=4\left(x-2\right)$$

Use the distributive property to multiply 4 by x-2.

$$x^{2}-x+\frac{26+4x^{2}-20x}{4}-2=4x-8$$

Divide each term of 26+4x^{2}-20x by 4 to get \frac{13}{2}+x^{2}-5x.

$$x^{2}-x+\frac{13}{2}+x^{2}-5x-2=4x-8$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-x+\frac{13}{2}-5x-2=4x-8$$

Combine -x and -5x to get -6x.

$$2x^{2}-6x+\frac{13}{2}-2=4x-8$$

Subtract 2 from \frac{13}{2} to get \frac{9}{2}.

$$2x^{2}-6x+\frac{9}{2}=4x-8$$

Subtract 4x from both sides.

$$2x^{2}-6x+\frac{9}{2}-4x=-8$$

Combine -6x and -4x to get -10x.

$$2x^{2}-10x+\frac{9}{2}=-8$$

Add 8 to both sides.

$$2x^{2}-10x+\frac{9}{2}+8=0$$

Add \frac{9}{2} and 8 to get \frac{25}{2}.

$$2x^{2}-10x+\frac{25}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -10 for b, and \frac{25}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 2\times \frac{25}{2}}}{2\times 2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 2\times \frac{25}{2}}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-10\right)±\sqrt{100-8\times \frac{25}{2}}}{2\times 2}$$

Multiply -8 times \frac{25}{2}.

$$x=\frac{-\left(-10\right)±\sqrt{100-100}}{2\times 2}$$

Add 100 to -100.

$$x=\frac{-\left(-10\right)±\sqrt{0}}{2\times 2}$$

Take the square root of 0.

$$x=-\frac{-10}{2\times 2}$$

The opposite of -10 is 10.

$$x=\frac{10}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{10}{4}$$

Reduce the fraction \frac{10}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$\left(x-\frac{1}{2}\right)^{2}-\left(\frac{2x-3}{2}-1\right)^{2}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{1}{2}\right)^{2}.

$$x^{2}-x+\frac{1}{4}-\left(\frac{2x-3}{2}-1\right)^{2}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{2x-3}{2}-1\right)^{2}.

$$x^{2}-x+\frac{1}{4}-\left(\left(\frac{2x-3}{2}\right)^{2}-2\times \frac{2x-3}{2}+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

To raise \frac{2x-3}{2} to a power, raise both numerator and denominator to the power and then divide.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}-2\times \frac{2x-3}{2}+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Express -2\times \frac{2x-3}{2} as a single fraction.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}+\frac{-2\left(2x-3\right)}{2}+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Cancel out 2 and 2.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}-\left(2x-3\right)+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

To find the opposite of 2x-3, find the opposite of each term.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}-2x+3+1\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Add 3 and 1 to get 4.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}-2x+4\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -2x+4 times \frac{2^{2}}{2^{2}}.

$$x^{2}-x+\frac{1}{4}-\left(\frac{\left(2x-3\right)^{2}}{2^{2}}+\frac{\left(-2x+4\right)\times 2^{2}}{2^{2}}\right)+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Since \frac{\left(2x-3\right)^{2}}{2^{2}} and \frac{\left(-2x+4\right)\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$x^{2}-x+\frac{1}{4}-\frac{\left(2x-3\right)^{2}+\left(-2x+4\right)\times 2^{2}}{2^{2}}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Do the multiplications in \left(2x-3\right)^{2}+\left(-2x+4\right)\times 2^{2}.

$$x^{2}-x+\frac{1}{4}-\frac{4x^{2}-12x+9-8x+16}{2^{2}}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Combine like terms in 4x^{2}-12x+9-8x+16.

$$x^{2}-x+\frac{1}{4}-\frac{4x^{2}-20x+25}{2^{2}}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}-x+\frac{1}{4}-\frac{4x^{2}-20x+25}{4}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Since \frac{1}{4} and \frac{4x^{2}-20x+25}{4} have the same denominator, add them by adding their numerators.

$$x^{2}-x+\frac{1+4x^{2}-20x+25}{4}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Combine like terms in 1+4x^{2}-20x+25.

$$x^{2}-x+\frac{26+4x^{2}-20x}{4}+6\left(\frac{2}{3}-1\right)=4\left(x-2\right)$$

Subtract 1 from \frac{2}{3} to get -\frac{1}{3}.

$$x^{2}-x+\frac{26+4x^{2}-20x}{4}+6\left(-\frac{1}{3}\right)=4\left(x-2\right)$$

Multiply 6 and -\frac{1}{3} to get -2.

$$x^{2}-x+\frac{26+4x^{2}-20x}{4}-2=4\left(x-2\right)$$

Use the distributive property to multiply 4 by x-2.

$$x^{2}-x+\frac{26+4x^{2}-20x}{4}-2=4x-8$$

Divide each term of 26+4x^{2}-20x by 4 to get \frac{13}{2}+x^{2}-5x.

$$x^{2}-x+\frac{13}{2}+x^{2}-5x-2=4x-8$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-x+\frac{13}{2}-5x-2=4x-8$$

Combine -x and -5x to get -6x.

$$2x^{2}-6x+\frac{13}{2}-2=4x-8$$

Subtract 2 from \frac{13}{2} to get \frac{9}{2}.

$$2x^{2}-6x+\frac{9}{2}=4x-8$$

Subtract 4x from both sides.

$$2x^{2}-6x+\frac{9}{2}-4x=-8$$

Combine -6x and -4x to get -10x.

$$2x^{2}-10x+\frac{9}{2}=-8$$

Subtract \frac{9}{2} from both sides.

$$2x^{2}-10x=-8-\frac{9}{2}$$

Subtract \frac{9}{2} from -8 to get -\frac{25}{2}.

$$2x^{2}-10x=-\frac{25}{2}$$

Divide both sides by 2.

$$\frac{2x^{2}-10x}{2}=-\frac{\frac{25}{2}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{10}{2}\right)x=-\frac{\frac{25}{2}}{2}$$

Divide -10 by 2.

$$x^{2}-5x=-\frac{\frac{25}{2}}{2}$$

Divide -\frac{25}{2} by 2.

$$x^{2}-5x=-\frac{25}{4}$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-\frac{25}{4}+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=\frac{-25+25}{4}$$

Add -\frac{25}{4} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-5x+\frac{25}{4}=0$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{5}{2}=0 x-\frac{5}{2}=0$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{5}{2} x=\frac{5}{2}$$

The equation is now solved. Solutions are the same.

$$x=\frac{5}{2}$$