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

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

Solve for x

$$x=\frac{2}{7}\approx 0.285714286$$

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

Steps Using Factoring By Grouping

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

$$2\left(2x-1\right)^{2}-\left(x-2\right)\left(1-2x\right)=6\left(1-2x\right)^{2}$$

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

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

Use the distributive property to multiply 2 by 4x^{2}-4x+1.

$$8x^{2}-8x+2-\left(x-2\right)\left(1-2x\right)=6\left(1-2x\right)^{2}$$

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

$$8x^{2}-8x+2-\left(5x-2x^{2}-2\right)=6\left(1-2x\right)^{2}$$

To find the opposite of 5x-2x^{2}-2, find the opposite of each term.

$$8x^{2}-8x+2-5x+2x^{2}+2=6\left(1-2x\right)^{2}$$

Combine -8x and -5x to get -13x.

$$8x^{2}-13x+2+2x^{2}+2=6\left(1-2x\right)^{2}$$

Combine 8x^{2} and 2x^{2} to get 10x^{2}.

$$10x^{2}-13x+2+2=6\left(1-2x\right)^{2}$$

Add 2 and 2 to get 4.

$$10x^{2}-13x+4=6\left(1-2x\right)^{2}$$

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

$$10x^{2}-13x+4=6\left(1-4x+4x^{2}\right)$$

Use the distributive property to multiply 6 by 1-4x+4x^{2}.

$$10x^{2}-13x+4=6-24x+24x^{2}$$

Subtract 6 from both sides.

$$10x^{2}-13x+4-6=-24x+24x^{2}$$

Subtract 6 from 4 to get -2.

$$10x^{2}-13x-2=-24x+24x^{2}$$

Add 24x to both sides.

$$10x^{2}-13x-2+24x=24x^{2}$$

Combine -13x and 24x to get 11x.

$$10x^{2}+11x-2=24x^{2}$$

Subtract 24x^{2} from both sides.

$$10x^{2}+11x-2-24x^{2}=0$$

Combine 10x^{2} and -24x^{2} to get -14x^{2}.

$$-14x^{2}+11x-2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -14x^{2}+ax+bx-2. To find a and b, set up a system to be solved.

$$a+b=11 ab=-14\left(-2\right)=28$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 28.

$$1,28 2,14 4,7$$

Calculate the sum for each pair.

$$1+28=29 2+14=16 4+7=11$$

The solution is the pair that gives sum 11.

$$a=7 b=4$$

Rewrite -14x^{2}+11x-2 as \left(-14x^{2}+7x\right)+\left(4x-2\right).

$$\left(-14x^{2}+7x\right)+\left(4x-2\right)$$

Factor out -7x in the first and 2 in the second group.

$$-7x\left(2x-1\right)+2\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(-7x+2\right)$$

To find equation solutions, solve 2x-1=0 and -7x+2=0.

$$x=\frac{1}{2} x=\frac{2}{7}$$

Steps Using the Quadratic Formula

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

$$2\left(2x-1\right)^{2}-\left(x-2\right)\left(1-2x\right)=6\left(1-2x\right)^{2}$$

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

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

Use the distributive property to multiply 2 by 4x^{2}-4x+1.

$$8x^{2}-8x+2-\left(x-2\right)\left(1-2x\right)=6\left(1-2x\right)^{2}$$

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

$$8x^{2}-8x+2-\left(5x-2x^{2}-2\right)=6\left(1-2x\right)^{2}$$

To find the opposite of 5x-2x^{2}-2, find the opposite of each term.

$$8x^{2}-8x+2-5x+2x^{2}+2=6\left(1-2x\right)^{2}$$

Combine -8x and -5x to get -13x.

$$8x^{2}-13x+2+2x^{2}+2=6\left(1-2x\right)^{2}$$

Combine 8x^{2} and 2x^{2} to get 10x^{2}.

$$10x^{2}-13x+2+2=6\left(1-2x\right)^{2}$$

Add 2 and 2 to get 4.

$$10x^{2}-13x+4=6\left(1-2x\right)^{2}$$

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

$$10x^{2}-13x+4=6\left(1-4x+4x^{2}\right)$$

Use the distributive property to multiply 6 by 1-4x+4x^{2}.

$$10x^{2}-13x+4=6-24x+24x^{2}$$

Subtract 6 from both sides.

$$10x^{2}-13x+4-6=-24x+24x^{2}$$

Subtract 6 from 4 to get -2.

$$10x^{2}-13x-2=-24x+24x^{2}$$

Add 24x to both sides.

$$10x^{2}-13x-2+24x=24x^{2}$$

Combine -13x and 24x to get 11x.

$$10x^{2}+11x-2=24x^{2}$$

Subtract 24x^{2} from both sides.

$$10x^{2}+11x-2-24x^{2}=0$$

Combine 10x^{2} and -24x^{2} to get -14x^{2}.

$$-14x^{2}+11x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -14 for a, 11 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-14\right)\left(-2\right)}}{2\left(-14\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-14\right)\left(-2\right)}}{2\left(-14\right)}$$

Multiply -4 times -14.

$$x=\frac{-11±\sqrt{121+56\left(-2\right)}}{2\left(-14\right)}$$

Multiply 56 times -2.

$$x=\frac{-11±\sqrt{121-112}}{2\left(-14\right)}$$

Add 121 to -112.

$$x=\frac{-11±\sqrt{9}}{2\left(-14\right)}$$

Take the square root of 9.

$$x=\frac{-11±3}{2\left(-14\right)}$$

Multiply 2 times -14.

$$x=\frac{-11±3}{-28}$$

Now solve the equation x=\frac{-11±3}{-28} when ± is plus. Add -11 to 3.

$$x=-\frac{8}{-28}$$

Reduce the fraction \frac{-8}{-28} to lowest terms by extracting and canceling out 4.

$$x=\frac{2}{7}$$

Now solve the equation x=\frac{-11±3}{-28} when ± is minus. Subtract 3 from -11.

$$x=-\frac{14}{-28}$$

Reduce the fraction \frac{-14}{-28} to lowest terms by extracting and canceling out 14.

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

The equation is now solved.

$$x=\frac{2}{7} x=\frac{1}{2}$$

Steps for Completing the Square

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

$$2\left(2x-1\right)^{2}-\left(x-2\right)\left(1-2x\right)=6\left(1-2x\right)^{2}$$

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

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

Use the distributive property to multiply 2 by 4x^{2}-4x+1.

$$8x^{2}-8x+2-\left(x-2\right)\left(1-2x\right)=6\left(1-2x\right)^{2}$$

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

$$8x^{2}-8x+2-\left(5x-2x^{2}-2\right)=6\left(1-2x\right)^{2}$$

To find the opposite of 5x-2x^{2}-2, find the opposite of each term.

$$8x^{2}-8x+2-5x+2x^{2}+2=6\left(1-2x\right)^{2}$$

Combine -8x and -5x to get -13x.

$$8x^{2}-13x+2+2x^{2}+2=6\left(1-2x\right)^{2}$$

Combine 8x^{2} and 2x^{2} to get 10x^{2}.

$$10x^{2}-13x+2+2=6\left(1-2x\right)^{2}$$

Add 2 and 2 to get 4.

$$10x^{2}-13x+4=6\left(1-2x\right)^{2}$$

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

$$10x^{2}-13x+4=6\left(1-4x+4x^{2}\right)$$

Use the distributive property to multiply 6 by 1-4x+4x^{2}.

$$10x^{2}-13x+4=6-24x+24x^{2}$$

Add 24x to both sides.

$$10x^{2}-13x+4+24x=6+24x^{2}$$

Combine -13x and 24x to get 11x.

$$10x^{2}+11x+4=6+24x^{2}$$

Subtract 24x^{2} from both sides.

$$10x^{2}+11x+4-24x^{2}=6$$

Combine 10x^{2} and -24x^{2} to get -14x^{2}.

$$-14x^{2}+11x+4=6$$

Subtract 4 from both sides.

$$-14x^{2}+11x=6-4$$

Subtract 4 from 6 to get 2.

$$-14x^{2}+11x=2$$

Divide both sides by -14.

$$\frac{-14x^{2}+11x}{-14}=\frac{2}{-14}$$

Dividing by -14 undoes the multiplication by -14.

$$x^{2}+\frac{11}{-14}x=\frac{2}{-14}$$

Divide 11 by -14.

$$x^{2}-\frac{11}{14}x=\frac{2}{-14}$$

Reduce the fraction \frac{2}{-14} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{11}{14}x=-\frac{1}{7}$$

Divide -\frac{11}{14}, the coefficient of the x term, by 2 to get -\frac{11}{28}. Then add the square of -\frac{11}{28} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{14}x+\left(-\frac{11}{28}\right)^{2}=-\frac{1}{7}+\left(-\frac{11}{28}\right)^{2}$$

Square -\frac{11}{28} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{14}x+\frac{121}{784}=-\frac{1}{7}+\frac{121}{784}$$

Add -\frac{1}{7} to \frac{121}{784} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{14}x+\frac{121}{784}=\frac{9}{784}$$

Factor x^{2}-\frac{11}{14}x+\frac{121}{784}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{28}\right)^{2}=\frac{9}{784}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{28}\right)^{2}}=\sqrt{\frac{9}{784}}$$

Simplify.

$$x-\frac{11}{28}=\frac{3}{28} x-\frac{11}{28}=-\frac{3}{28}$$

Add \frac{11}{28} to both sides of the equation.

$$x=\frac{1}{2} x=\frac{2}{7}$$