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

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

Solve for x

$$x=-3$$

$$x=\frac{9}{10}=0.9$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 12, the least common multiple of 4,3,2,6.

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

Use the distributive property to multiply 3 by x-1.

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

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

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 2 is 6. Multiply \frac{2x-3}{3} times \frac{2}{2}. Multiply \frac{x-1}{2} times \frac{3}{3}.

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

Since \frac{2\left(2x-3\right)}{6} and \frac{3\left(x-1\right)}{6} have the same denominator, subtract them by subtracting their numerators.

$$3x^{2}-3=12\left(\frac{1}{3}x+\frac{x-2}{2}\right)\times \frac{2\left(2x-3\right)-3\left(x-1\right)}{6}-2x^{2}$$

Do the multiplications in 2\left(2x-3\right)-3\left(x-1\right).

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

Combine like terms in 4x-6-3x+3.

$$3x^{2}-3=12\left(\frac{1}{3}x+\frac{x-2}{2}\right)\times \frac{x-3}{6}-2x^{2}$$

Cancel out 6, the greatest common factor in 12 and 6.

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

Use the distributive property to multiply 2\left(x-3\right) by \frac{1}{3}x+\frac{x-2}{2} and combine like terms.

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

Use the distributive property to multiply \frac{2}{3} by x-3.

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

Use the distributive property to multiply \frac{2}{3}x-2 by x.

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

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

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

Cancel out 2 and 2.

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

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

$$3x^{2}-3=\frac{2}{3}x^{2}-2x+x^{2}-5x+6-2x^{2}$$

Combine \frac{2}{3}x^{2} and x^{2} to get \frac{5}{3}x^{2}.

$$3x^{2}-3=\frac{5}{3}x^{2}-2x-5x+6-2x^{2}$$

Combine -2x and -5x to get -7x.

$$3x^{2}-3=\frac{5}{3}x^{2}-7x+6-2x^{2}$$

Combine \frac{5}{3}x^{2} and -2x^{2} to get -\frac{1}{3}x^{2}.

$$3x^{2}-3=-\frac{1}{3}x^{2}-7x+6$$

Add \frac{1}{3}x^{2} to both sides.

$$3x^{2}-3+\frac{1}{3}x^{2}=-7x+6$$

Combine 3x^{2} and \frac{1}{3}x^{2} to get \frac{10}{3}x^{2}.

$$\frac{10}{3}x^{2}-3=-7x+6$$

Add 7x to both sides.

$$\frac{10}{3}x^{2}-3+7x=6$$

Subtract 6 from both sides.

$$\frac{10}{3}x^{2}-3+7x-6=0$$

Subtract 6 from -3 to get -9.

$$\frac{10}{3}x^{2}-9+7x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$\frac{10}{3}x^{2}+7x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{10}{3} for a, 7 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times \frac{10}{3}\left(-9\right)}}{2\times \frac{10}{3}}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times \frac{10}{3}\left(-9\right)}}{2\times \frac{10}{3}}$$

Multiply -4 times \frac{10}{3}.

$$x=\frac{-7±\sqrt{49-\frac{40}{3}\left(-9\right)}}{2\times \frac{10}{3}}$$

Multiply -\frac{40}{3} times -9.

$$x=\frac{-7±\sqrt{49+120}}{2\times \frac{10}{3}}$$

Add 49 to 120.

$$x=\frac{-7±\sqrt{169}}{2\times \frac{10}{3}}$$

Take the square root of 169.

$$x=\frac{-7±13}{2\times \frac{10}{3}}$$

Multiply 2 times \frac{10}{3}.

$$x=\frac{-7±13}{\frac{20}{3}}$$

Now solve the equation x=\frac{-7±13}{\frac{20}{3}} when ± is plus. Add -7 to 13.

$$x=\frac{6}{\frac{20}{3}}$$

Divide 6 by \frac{20}{3} by multiplying 6 by the reciprocal of \frac{20}{3}.

$$x=\frac{9}{10}$$

Now solve the equation x=\frac{-7±13}{\frac{20}{3}} when ± is minus. Subtract 13 from -7.

$$x=-\frac{20}{\frac{20}{3}}$$

Divide -20 by \frac{20}{3} by multiplying -20 by the reciprocal of \frac{20}{3}.

$$x=-3$$

The equation is now solved.

$$x=\frac{9}{10} x=-3$$

Steps for Completing the Square

Multiply both sides of the equation by 12, the least common multiple of 4,3,2,6.

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

Use the distributive property to multiply 3 by x-1.

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

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

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 2 is 6. Multiply \frac{2x-3}{3} times \frac{2}{2}. Multiply \frac{x-1}{2} times \frac{3}{3}.

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

Since \frac{2\left(2x-3\right)}{6} and \frac{3\left(x-1\right)}{6} have the same denominator, subtract them by subtracting their numerators.

$$3x^{2}-3=12\left(\frac{1}{3}x+\frac{x-2}{2}\right)\times \frac{2\left(2x-3\right)-3\left(x-1\right)}{6}-2x^{2}$$

Do the multiplications in 2\left(2x-3\right)-3\left(x-1\right).

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

Combine like terms in 4x-6-3x+3.

$$3x^{2}-3=12\left(\frac{1}{3}x+\frac{x-2}{2}\right)\times \frac{x-3}{6}-2x^{2}$$

Cancel out 6, the greatest common factor in 12 and 6.

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

Use the distributive property to multiply 2\left(x-3\right) by \frac{1}{3}x+\frac{x-2}{2} and combine like terms.

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

Use the distributive property to multiply \frac{2}{3} by x-3.

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

Use the distributive property to multiply \frac{2}{3}x-2 by x.

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

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

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

Cancel out 2 and 2.

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

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

$$3x^{2}-3=\frac{2}{3}x^{2}-2x+x^{2}-5x+6-2x^{2}$$

Combine \frac{2}{3}x^{2} and x^{2} to get \frac{5}{3}x^{2}.

$$3x^{2}-3=\frac{5}{3}x^{2}-2x-5x+6-2x^{2}$$

Combine -2x and -5x to get -7x.

$$3x^{2}-3=\frac{5}{3}x^{2}-7x+6-2x^{2}$$

Combine \frac{5}{3}x^{2} and -2x^{2} to get -\frac{1}{3}x^{2}.

$$3x^{2}-3=-\frac{1}{3}x^{2}-7x+6$$

Add \frac{1}{3}x^{2} to both sides.

$$3x^{2}-3+\frac{1}{3}x^{2}=-7x+6$$

Combine 3x^{2} and \frac{1}{3}x^{2} to get \frac{10}{3}x^{2}.

$$\frac{10}{3}x^{2}-3=-7x+6$$

Add 7x to both sides.

$$\frac{10}{3}x^{2}-3+7x=6$$

Add 3 to both sides.

$$\frac{10}{3}x^{2}+7x=6+3$$

Add 6 and 3 to get 9.

$$\frac{10}{3}x^{2}+7x=9$$

Divide both sides of the equation by \frac{10}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{10}{3}x^{2}+7x}{\frac{10}{3}}=\frac{9}{\frac{10}{3}}$$

Dividing by \frac{10}{3} undoes the multiplication by \frac{10}{3}.

$$x^{2}+\frac{7}{\frac{10}{3}}x=\frac{9}{\frac{10}{3}}$$

Divide 7 by \frac{10}{3} by multiplying 7 by the reciprocal of \frac{10}{3}.

$$x^{2}+\frac{21}{10}x=\frac{9}{\frac{10}{3}}$$

Divide 9 by \frac{10}{3} by multiplying 9 by the reciprocal of \frac{10}{3}.

$$x^{2}+\frac{21}{10}x=\frac{27}{10}$$

Divide \frac{21}{10}, the coefficient of the x term, by 2 to get \frac{21}{20}. Then add the square of \frac{21}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{21}{10}x+\left(\frac{21}{20}\right)^{2}=\frac{27}{10}+\left(\frac{21}{20}\right)^{2}$$

Square \frac{21}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{21}{10}x+\frac{441}{400}=\frac{27}{10}+\frac{441}{400}$$

Add \frac{27}{10} to \frac{441}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{21}{10}x+\frac{441}{400}=\frac{1521}{400}$$

Factor x^{2}+\frac{21}{10}x+\frac{441}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{21}{20}\right)^{2}=\frac{1521}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{21}{20}\right)^{2}}=\sqrt{\frac{1521}{400}}$$

Simplify.

$$x+\frac{21}{20}=\frac{39}{20} x+\frac{21}{20}=-\frac{39}{20}$$

Subtract \frac{21}{20} from both sides of the equation.

$$x=\frac{9}{10} x=-3$$