Solve x^2-3x+5/4-2=(x+1/2)(x-3)+(x/2-1)^2

$$\frac { x ^ { 2 } - 3 x + 5 } { 4 } - 2 = ( x + \frac { 1 } { 2 } ) ( x - 3 ) + ( \frac { x } { 2 } - 1 ) ^ { 2 }$$

Solve for x

$$x = \frac{\sqrt{105} + 11}{8} \approx 2.655868846$$

$$x=\frac{11-\sqrt{105}}{8}\approx 0.094131154$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4, the least common multiple of 4,2.

$$x^{2}-3x+5-8=4\left(x+\frac{1}{2}\right)\left(x-3\right)+4\left(\frac{x}{2}-1\right)^{2}$$

Subtract 8 from 5 to get -3.

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

Use the distributive property to multiply 4 by x+\frac{1}{2}.

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

Use the distributive property to multiply 4x+2 by x-3 and combine like terms.

$$x^{2}-3x-3=4x^{2}-10x-6+4\left(\frac{x}{2}-1\right)^{2}$$

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

$$x^{2}-3x-3=4x^{2}-10x-6+4\left(\left(\frac{x}{2}\right)^{2}-2\times \frac{x}{2}+1\right)$$

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

$$x^{2}-3x-3=4x^{2}-10x-6+4\left(\frac{x^{2}}{2^{2}}-2\times \frac{x}{2}+1\right)$$

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

$$x^{2}-3x-3=4x^{2}-10x-6+4\left(\frac{x^{2}}{2^{2}}+\frac{-2x}{2}+1\right)$$

Cancel out 2 and 2.

$$x^{2}-3x-3=4x^{2}-10x-6+4\left(\frac{x^{2}}{2^{2}}-x+1\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -x+1 times \frac{2^{2}}{2^{2}}.

$$x^{2}-3x-3=4x^{2}-10x-6+4\left(\frac{x^{2}}{2^{2}}+\frac{\left(-x+1\right)\times 2^{2}}{2^{2}}\right)$$

Since \frac{x^{2}}{2^{2}} and \frac{\left(-x+1\right)\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$x^{2}-3x-3=4x^{2}-10x-6+4\times \frac{x^{2}+\left(-x+1\right)\times 2^{2}}{2^{2}}$$

Do the multiplications in x^{2}+\left(-x+1\right)\times 2^{2}.

$$x^{2}-3x-3=4x^{2}-10x-6+4\times \frac{x^{2}-4x+4}{2^{2}}$$

Express 4\times \frac{x^{2}-4x+4}{2^{2}} as a single fraction.

$$x^{2}-3x-3=4x^{2}-10x-6+\frac{4\left(x^{2}-4x+4\right)}{2^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4x^{2}-10x-6 times \frac{2^{2}}{2^{2}}.

$$x^{2}-3x-3=\frac{\left(4x^{2}-10x-6\right)\times 2^{2}}{2^{2}}+\frac{4\left(x^{2}-4x+4\right)}{2^{2}}$$

Since \frac{\left(4x^{2}-10x-6\right)\times 2^{2}}{2^{2}} and \frac{4\left(x^{2}-4x+4\right)}{2^{2}} have the same denominator, add them by adding their numerators.

$$x^{2}-3x-3=\frac{\left(4x^{2}-10x-6\right)\times 2^{2}+4\left(x^{2}-4x+4\right)}{2^{2}}$$

Do the multiplications in \left(4x^{2}-10x-6\right)\times 2^{2}+4\left(x^{2}-4x+4\right).

$$x^{2}-3x-3=\frac{16x^{2}-40x-24+4x^{2}-16x+16}{2^{2}}$$

Combine like terms in 16x^{2}-40x-24+4x^{2}-16x+16.

$$x^{2}-3x-3=\frac{20x^{2}-56x-8}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}-3x-3=\frac{20x^{2}-56x-8}{4}$$

Divide each term of 20x^{2}-56x-8 by 4 to get -2-14x+5x^{2}.

$$x^{2}-3x-3=-2-14x+5x^{2}$$

Subtract -2 from both sides.

$$x^{2}-3x-3-\left(-2\right)=-14x+5x^{2}$$

The opposite of -2 is 2.

$$x^{2}-3x-3+2=-14x+5x^{2}$$

Add 14x to both sides.

$$x^{2}-3x-3+2+14x=5x^{2}$$

Add -3 and 2 to get -1.

$$x^{2}-3x-1+14x=5x^{2}$$

Combine -3x and 14x to get 11x.

$$x^{2}+11x-1=5x^{2}$$

Subtract 5x^{2} from both sides.

$$x^{2}+11x-1-5x^{2}=0$$

Combine x^{2} and -5x^{2} to get -4x^{2}.

$$-4x^{2}+11x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 11 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-4\right)\left(-1\right)}}{2\left(-4\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-4\right)\left(-1\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-11±\sqrt{121+16\left(-1\right)}}{2\left(-4\right)}$$

Multiply 16 times -1.

$$x=\frac{-11±\sqrt{121-16}}{2\left(-4\right)}$$

Add 121 to -16.

$$x=\frac{-11±\sqrt{105}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-11±\sqrt{105}}{-8}$$

Now solve the equation x=\frac{-11±\sqrt{105}}{-8} when ± is plus. Add -11 to \sqrt{105}.

$$x=\frac{\sqrt{105}-11}{-8}$$

Divide -11+\sqrt{105} by -8.

$$x=\frac{11-\sqrt{105}}{8}$$

Now solve the equation x=\frac{-11±\sqrt{105}}{-8} when ± is minus. Subtract \sqrt{105} from -11.

$$x=\frac{-\sqrt{105}-11}{-8}$$

Divide -11-\sqrt{105} by -8.

$$x=\frac{\sqrt{105}+11}{8}$$

The equation is now solved.

$$x=\frac{11-\sqrt{105}}{8} x=\frac{\sqrt{105}+11}{8}$$

Steps for Completing the Square

Multiply both sides of the equation by 4, the least common multiple of 4,2.

$$x^{2}-3x+5-8=4\left(x+\frac{1}{2}\right)\left(x-3\right)+4\left(\frac{x}{2}-1\right)^{2}$$

Subtract 8 from 5 to get -3.

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

Use the distributive property to multiply 4 by x+\frac{1}{2}.

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

Use the distributive property to multiply 4x+2 by x-3 and combine like terms.

$$x^{2}-3x-3=4x^{2}-10x-6+4\left(\frac{x}{2}-1\right)^{2}$$

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

$$x^{2}-3x-3=4x^{2}-10x-6+4\left(\left(\frac{x}{2}\right)^{2}-2\times \frac{x}{2}+1\right)$$

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

$$x^{2}-3x-3=4x^{2}-10x-6+4\left(\frac{x^{2}}{2^{2}}-2\times \frac{x}{2}+1\right)$$

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

$$x^{2}-3x-3=4x^{2}-10x-6+4\left(\frac{x^{2}}{2^{2}}+\frac{-2x}{2}+1\right)$$

Cancel out 2 and 2.

$$x^{2}-3x-3=4x^{2}-10x-6+4\left(\frac{x^{2}}{2^{2}}-x+1\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -x+1 times \frac{2^{2}}{2^{2}}.

$$x^{2}-3x-3=4x^{2}-10x-6+4\left(\frac{x^{2}}{2^{2}}+\frac{\left(-x+1\right)\times 2^{2}}{2^{2}}\right)$$

Since \frac{x^{2}}{2^{2}} and \frac{\left(-x+1\right)\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$x^{2}-3x-3=4x^{2}-10x-6+4\times \frac{x^{2}+\left(-x+1\right)\times 2^{2}}{2^{2}}$$

Do the multiplications in x^{2}+\left(-x+1\right)\times 2^{2}.

$$x^{2}-3x-3=4x^{2}-10x-6+4\times \frac{x^{2}-4x+4}{2^{2}}$$

Express 4\times \frac{x^{2}-4x+4}{2^{2}} as a single fraction.

$$x^{2}-3x-3=4x^{2}-10x-6+\frac{4\left(x^{2}-4x+4\right)}{2^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4x^{2}-10x-6 times \frac{2^{2}}{2^{2}}.

$$x^{2}-3x-3=\frac{\left(4x^{2}-10x-6\right)\times 2^{2}}{2^{2}}+\frac{4\left(x^{2}-4x+4\right)}{2^{2}}$$

Since \frac{\left(4x^{2}-10x-6\right)\times 2^{2}}{2^{2}} and \frac{4\left(x^{2}-4x+4\right)}{2^{2}} have the same denominator, add them by adding their numerators.

$$x^{2}-3x-3=\frac{\left(4x^{2}-10x-6\right)\times 2^{2}+4\left(x^{2}-4x+4\right)}{2^{2}}$$

Do the multiplications in \left(4x^{2}-10x-6\right)\times 2^{2}+4\left(x^{2}-4x+4\right).

$$x^{2}-3x-3=\frac{16x^{2}-40x-24+4x^{2}-16x+16}{2^{2}}$$

Combine like terms in 16x^{2}-40x-24+4x^{2}-16x+16.

$$x^{2}-3x-3=\frac{20x^{2}-56x-8}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}-3x-3=\frac{20x^{2}-56x-8}{4}$$

Divide each term of 20x^{2}-56x-8 by 4 to get -2-14x+5x^{2}.

$$x^{2}-3x-3=-2-14x+5x^{2}$$

Add 14x to both sides.

$$x^{2}-3x-3+14x=-2+5x^{2}$$

Combine -3x and 14x to get 11x.

$$x^{2}+11x-3=-2+5x^{2}$$

Subtract 5x^{2} from both sides.

$$x^{2}+11x-3-5x^{2}=-2$$

Combine x^{2} and -5x^{2} to get -4x^{2}.

$$-4x^{2}+11x-3=-2$$

Add 3 to both sides.

$$-4x^{2}+11x=-2+3$$

Add -2 and 3 to get 1.

$$-4x^{2}+11x=1$$

Divide both sides by -4.

$$\frac{-4x^{2}+11x}{-4}=\frac{1}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{11}{-4}x=\frac{1}{-4}$$

Divide 11 by -4.

$$x^{2}-\frac{11}{4}x=\frac{1}{-4}$$

Divide 1 by -4.

$$x^{2}-\frac{11}{4}x=-\frac{1}{4}$$

Divide -\frac{11}{4}, the coefficient of the x term, by 2 to get -\frac{11}{8}. Then add the square of -\frac{11}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{4}x+\left(-\frac{11}{8}\right)^{2}=-\frac{1}{4}+\left(-\frac{11}{8}\right)^{2}$$

Square -\frac{11}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{4}x+\frac{121}{64}=-\frac{1}{4}+\frac{121}{64}$$

Add -\frac{1}{4} to \frac{121}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{4}x+\frac{121}{64}=\frac{105}{64}$$

Factor x^{2}-\frac{11}{4}x+\frac{121}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{8}\right)^{2}=\frac{105}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{8}\right)^{2}}=\sqrt{\frac{105}{64}}$$

Simplify.

$$x-\frac{11}{8}=\frac{\sqrt{105}}{8} x-\frac{11}{8}=-\frac{\sqrt{105}}{8}$$

Add \frac{11}{8} to both sides of the equation.

$$x=\frac{\sqrt{105}+11}{8} x=\frac{11-\sqrt{105}}{8}$$