Solve {l}{(x/2-1)^2+2(2x-1)(2x+1)/3+11/12=(x+2)(3x+1)/4-frac{x(3-4x)}{6}}{text{uazione.}}{qquad2(4x-1)-3(x^2-4)-9=6(2+x)-(x+5)^2}

$$\left. \begin{array} { l } { ( \frac { x } { 2 } - 1 ) ^ { 2 } + \frac { 2 ( 2 x - 1 ) ( 2 x + 1 ) } { 3 } + \frac { 11 } { 12 } = \frac { ( x + 2 ) ( 3 x + 1 ) } { 4 } - \frac { x ( 3 - 4 x ) } { 6 } } \\ { \text { uazione. } } \\ { \qquad 2 ( 4 x - 1 ) - 3 ( x ^ { 2 } - 4 ) - 9 = 6 ( 2 + x ) - ( x + 5 ) ^ { 2 } } \end{array} \right.$$

Solve for x

$$x=\frac{1}{2}=0.5$$

$$x=1$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 12, the least common multiple of 3,12,4,6.

$$12\left(\frac{x}{2}-1\right)^{2}+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

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

$$12\left(\left(\frac{x}{2}\right)^{2}-2\times \frac{x}{2}+1\right)+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

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

$$12\left(\frac{x^{2}}{2^{2}}-2\times \frac{x}{2}+1\right)+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

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

$$12\left(\frac{x^{2}}{2^{2}}+\frac{-2x}{2}+1\right)+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Cancel out 2 and 2.

$$12\left(\frac{x^{2}}{2^{2}}-x+1\right)+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

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

$$12\left(\frac{x^{2}}{2^{2}}+\frac{\left(-x+1\right)\times 2^{2}}{2^{2}}\right)+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

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.

$$12\times \frac{x^{2}+\left(-x+1\right)\times 2^{2}}{2^{2}}+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

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

$$12\times \frac{x^{2}-4x+4}{2^{2}}+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Express 12\times \frac{x^{2}-4x+4}{2^{2}} as a single fraction.

$$\frac{12\left(x^{2}-4x+4\right)}{2^{2}}+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Multiply 4 and 2 to get 8.

$$\frac{12\left(x^{2}-4x+4\right)}{2^{2}}+8\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Use the distributive property to multiply 8 by 2x-1.

$$\frac{12\left(x^{2}-4x+4\right)}{2^{2}}+\left(16x-8\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Use the distributive property to multiply 16x-8 by 2x+1 and combine like terms.

$$\frac{12\left(x^{2}-4x+4\right)}{2^{2}}+32x^{2}-8+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Add -8 and 11 to get 3.

$$\frac{12\left(x^{2}-4x+4\right)}{2^{2}}+32x^{2}+3=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 32x^{2}+3 times \frac{2^{2}}{2^{2}}.

$$\frac{12\left(x^{2}-4x+4\right)}{2^{2}}+\frac{\left(32x^{2}+3\right)\times 2^{2}}{2^{2}}=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Since \frac{12\left(x^{2}-4x+4\right)}{2^{2}} and \frac{\left(32x^{2}+3\right)\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{12\left(x^{2}-4x+4\right)+\left(32x^{2}+3\right)\times 2^{2}}{2^{2}}=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Do the multiplications in 12\left(x^{2}-4x+4\right)+\left(32x^{2}+3\right)\times 2^{2}.

$$\frac{12x^{2}-48x+48+128x^{2}+12}{2^{2}}=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Combine like terms in 12x^{2}-48x+48+128x^{2}+12.

$$\frac{140x^{2}-48x+60}{2^{2}}=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Use the distributive property to multiply 3 by x+2.

$$\frac{140x^{2}-48x+60}{2^{2}}=\left(3x+6\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Use the distributive property to multiply 3x+6 by 3x+1 and combine like terms.

$$\frac{140x^{2}-48x+60}{2^{2}}=9x^{2}+21x+6-2x\left(3-4x\right)$$

Use the distributive property to multiply -2x by 3-4x.

$$\frac{140x^{2}-48x+60}{2^{2}}=9x^{2}+21x+6-6x+8x^{2}$$

Combine 21x and -6x to get 15x.

$$\frac{140x^{2}-48x+60}{2^{2}}=9x^{2}+15x+6+8x^{2}$$

Combine 9x^{2} and 8x^{2} to get 17x^{2}.

$$\frac{140x^{2}-48x+60}{2^{2}}=17x^{2}+15x+6$$

Calculate 2 to the power of 2 and get 4.

$$\frac{140x^{2}-48x+60}{4}=17x^{2}+15x+6$$

Divide each term of 140x^{2}-48x+60 by 4 to get 35x^{2}-12x+15.

$$35x^{2}-12x+15=17x^{2}+15x+6$$

Subtract 17x^{2} from both sides.

$$35x^{2}-12x+15-17x^{2}=15x+6$$

Combine 35x^{2} and -17x^{2} to get 18x^{2}.

$$18x^{2}-12x+15=15x+6$$

Subtract 15x from both sides.

$$18x^{2}-12x+15-15x=6$$

Combine -12x and -15x to get -27x.

$$18x^{2}-27x+15=6$$

Subtract 6 from both sides.

$$18x^{2}-27x+15-6=0$$

Subtract 6 from 15 to get 9.

$$18x^{2}-27x+9=0$$

Divide both sides by 9.

$$2x^{2}-3x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-3 ab=2\times 1=2$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-2 b=-1$$

Rewrite 2x^{2}-3x+1 as \left(2x^{2}-2x\right)+\left(-x+1\right).

$$\left(2x^{2}-2x\right)+\left(-x+1\right)$$

Factor out 2x in the first and -1 in the second group.

$$2x\left(x-1\right)-\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(2x-1\right)$$

To find equation solutions, solve x-1=0 and 2x-1=0.

$$x=1 x=\frac{1}{2}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 12, the least common multiple of 3,12,4,6.

$$12\left(\frac{x}{2}-1\right)^{2}+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

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

$$12\left(\left(\frac{x}{2}\right)^{2}-2\times \frac{x}{2}+1\right)+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

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

$$12\left(\frac{x^{2}}{2^{2}}-2\times \frac{x}{2}+1\right)+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

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

$$12\left(\frac{x^{2}}{2^{2}}+\frac{-2x}{2}+1\right)+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Cancel out 2 and 2.

$$12\left(\frac{x^{2}}{2^{2}}-x+1\right)+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

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

$$12\left(\frac{x^{2}}{2^{2}}+\frac{\left(-x+1\right)\times 2^{2}}{2^{2}}\right)+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

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.

$$12\times \frac{x^{2}+\left(-x+1\right)\times 2^{2}}{2^{2}}+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

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

$$12\times \frac{x^{2}-4x+4}{2^{2}}+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Express 12\times \frac{x^{2}-4x+4}{2^{2}} as a single fraction.

$$\frac{12\left(x^{2}-4x+4\right)}{2^{2}}+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Multiply 4 and 2 to get 8.

$$\frac{12\left(x^{2}-4x+4\right)}{2^{2}}+8\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Use the distributive property to multiply 8 by 2x-1.

$$\frac{12\left(x^{2}-4x+4\right)}{2^{2}}+\left(16x-8\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Use the distributive property to multiply 16x-8 by 2x+1 and combine like terms.

$$\frac{12\left(x^{2}-4x+4\right)}{2^{2}}+32x^{2}-8+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Add -8 and 11 to get 3.

$$\frac{12\left(x^{2}-4x+4\right)}{2^{2}}+32x^{2}+3=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 32x^{2}+3 times \frac{2^{2}}{2^{2}}.

$$\frac{12\left(x^{2}-4x+4\right)}{2^{2}}+\frac{\left(32x^{2}+3\right)\times 2^{2}}{2^{2}}=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Since \frac{12\left(x^{2}-4x+4\right)}{2^{2}} and \frac{\left(32x^{2}+3\right)\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{12\left(x^{2}-4x+4\right)+\left(32x^{2}+3\right)\times 2^{2}}{2^{2}}=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Do the multiplications in 12\left(x^{2}-4x+4\right)+\left(32x^{2}+3\right)\times 2^{2}.

$$\frac{12x^{2}-48x+48+128x^{2}+12}{2^{2}}=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Combine like terms in 12x^{2}-48x+48+128x^{2}+12.

$$\frac{140x^{2}-48x+60}{2^{2}}=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Use the distributive property to multiply 3 by x+2.

$$\frac{140x^{2}-48x+60}{2^{2}}=\left(3x+6\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Use the distributive property to multiply 3x+6 by 3x+1 and combine like terms.

$$\frac{140x^{2}-48x+60}{2^{2}}=9x^{2}+21x+6-2x\left(3-4x\right)$$

Use the distributive property to multiply -2x by 3-4x.

$$\frac{140x^{2}-48x+60}{2^{2}}=9x^{2}+21x+6-6x+8x^{2}$$

Combine 21x and -6x to get 15x.

$$\frac{140x^{2}-48x+60}{2^{2}}=9x^{2}+15x+6+8x^{2}$$

Combine 9x^{2} and 8x^{2} to get 17x^{2}.

$$\frac{140x^{2}-48x+60}{2^{2}}=17x^{2}+15x+6$$

Calculate 2 to the power of 2 and get 4.

$$\frac{140x^{2}-48x+60}{4}=17x^{2}+15x+6$$

Divide each term of 140x^{2}-48x+60 by 4 to get 35x^{2}-12x+15.

$$35x^{2}-12x+15=17x^{2}+15x+6$$

Subtract 17x^{2} from both sides.

$$35x^{2}-12x+15-17x^{2}=15x+6$$

Combine 35x^{2} and -17x^{2} to get 18x^{2}.

$$18x^{2}-12x+15=15x+6$$

Subtract 15x from both sides.

$$18x^{2}-12x+15-15x=6$$

Combine -12x and -15x to get -27x.

$$18x^{2}-27x+15=6$$

Subtract 6 from both sides.

$$18x^{2}-27x+15-6=0$$

Subtract 6 from 15 to get 9.

$$18x^{2}-27x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, -27 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-27\right)±\sqrt{\left(-27\right)^{2}-4\times 18\times 9}}{2\times 18}$$

Square -27.

$$x=\frac{-\left(-27\right)±\sqrt{729-4\times 18\times 9}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-\left(-27\right)±\sqrt{729-72\times 9}}{2\times 18}$$

Multiply -72 times 9.

$$x=\frac{-\left(-27\right)±\sqrt{729-648}}{2\times 18}$$

Add 729 to -648.

$$x=\frac{-\left(-27\right)±\sqrt{81}}{2\times 18}$$

Take the square root of 81.

$$x=\frac{-\left(-27\right)±9}{2\times 18}$$

The opposite of -27 is 27.

$$x=\frac{27±9}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{27±9}{36}$$

Now solve the equation x=\frac{27±9}{36} when ± is plus. Add 27 to 9.

$$x=\frac{36}{36}$$

Divide 36 by 36.

$$x=1$$

Now solve the equation x=\frac{27±9}{36} when ± is minus. Subtract 9 from 27.

$$x=\frac{18}{36}$$

Reduce the fraction \frac{18}{36} to lowest terms by extracting and canceling out 18.

$$x=\frac{1}{2}$$

The equation is now solved.

$$x=1 x=\frac{1}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 12, the least common multiple of 3,12,4,6.

$$12\left(\frac{x}{2}-1\right)^{2}+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

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

$$12\left(\left(\frac{x}{2}\right)^{2}-2\times \frac{x}{2}+1\right)+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

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

$$12\left(\frac{x^{2}}{2^{2}}-2\times \frac{x}{2}+1\right)+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

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

$$12\left(\frac{x^{2}}{2^{2}}+\frac{-2x}{2}+1\right)+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Cancel out 2 and 2.

$$12\left(\frac{x^{2}}{2^{2}}-x+1\right)+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

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

$$12\left(\frac{x^{2}}{2^{2}}+\frac{\left(-x+1\right)\times 2^{2}}{2^{2}}\right)+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

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.

$$12\times \frac{x^{2}+\left(-x+1\right)\times 2^{2}}{2^{2}}+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

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

$$12\times \frac{x^{2}-4x+4}{2^{2}}+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Express 12\times \frac{x^{2}-4x+4}{2^{2}} as a single fraction.

$$\frac{12\left(x^{2}-4x+4\right)}{2^{2}}+4\times 2\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Multiply 4 and 2 to get 8.

$$\frac{12\left(x^{2}-4x+4\right)}{2^{2}}+8\left(2x-1\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Use the distributive property to multiply 8 by 2x-1.

$$\frac{12\left(x^{2}-4x+4\right)}{2^{2}}+\left(16x-8\right)\left(2x+1\right)+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Use the distributive property to multiply 16x-8 by 2x+1 and combine like terms.

$$\frac{12\left(x^{2}-4x+4\right)}{2^{2}}+32x^{2}-8+11=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Add -8 and 11 to get 3.

$$\frac{12\left(x^{2}-4x+4\right)}{2^{2}}+32x^{2}+3=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 32x^{2}+3 times \frac{2^{2}}{2^{2}}.

$$\frac{12\left(x^{2}-4x+4\right)}{2^{2}}+\frac{\left(32x^{2}+3\right)\times 2^{2}}{2^{2}}=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Since \frac{12\left(x^{2}-4x+4\right)}{2^{2}} and \frac{\left(32x^{2}+3\right)\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{12\left(x^{2}-4x+4\right)+\left(32x^{2}+3\right)\times 2^{2}}{2^{2}}=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Do the multiplications in 12\left(x^{2}-4x+4\right)+\left(32x^{2}+3\right)\times 2^{2}.

$$\frac{12x^{2}-48x+48+128x^{2}+12}{2^{2}}=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Combine like terms in 12x^{2}-48x+48+128x^{2}+12.

$$\frac{140x^{2}-48x+60}{2^{2}}=3\left(x+2\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Use the distributive property to multiply 3 by x+2.

$$\frac{140x^{2}-48x+60}{2^{2}}=\left(3x+6\right)\left(3x+1\right)-2x\left(3-4x\right)$$

Use the distributive property to multiply 3x+6 by 3x+1 and combine like terms.

$$\frac{140x^{2}-48x+60}{2^{2}}=9x^{2}+21x+6-2x\left(3-4x\right)$$

Use the distributive property to multiply -2x by 3-4x.

$$\frac{140x^{2}-48x+60}{2^{2}}=9x^{2}+21x+6-6x+8x^{2}$$

Combine 21x and -6x to get 15x.

$$\frac{140x^{2}-48x+60}{2^{2}}=9x^{2}+15x+6+8x^{2}$$

Combine 9x^{2} and 8x^{2} to get 17x^{2}.

$$\frac{140x^{2}-48x+60}{2^{2}}=17x^{2}+15x+6$$

Calculate 2 to the power of 2 and get 4.

$$\frac{140x^{2}-48x+60}{4}=17x^{2}+15x+6$$

Divide each term of 140x^{2}-48x+60 by 4 to get 35x^{2}-12x+15.

$$35x^{2}-12x+15=17x^{2}+15x+6$$

Subtract 17x^{2} from both sides.

$$35x^{2}-12x+15-17x^{2}=15x+6$$

Combine 35x^{2} and -17x^{2} to get 18x^{2}.

$$18x^{2}-12x+15=15x+6$$

Subtract 15x from both sides.

$$18x^{2}-12x+15-15x=6$$

Combine -12x and -15x to get -27x.

$$18x^{2}-27x+15=6$$

Subtract 15 from both sides.

$$18x^{2}-27x=6-15$$

Subtract 15 from 6 to get -9.

$$18x^{2}-27x=-9$$

Divide both sides by 18.

$$\frac{18x^{2}-27x}{18}=-\frac{9}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$x^{2}+\left(-\frac{27}{18}\right)x=-\frac{9}{18}$$

Reduce the fraction \frac{-27}{18} to lowest terms by extracting and canceling out 9.

$$x^{2}-\frac{3}{2}x=-\frac{9}{18}$$

Reduce the fraction \frac{-9}{18} to lowest terms by extracting and canceling out 9.

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

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{1}{2}+\frac{9}{16}$$

Add -\frac{1}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{1}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{1}{4} x-\frac{3}{4}=-\frac{1}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=1 x=\frac{1}{2}$$