Solve (1-2x)(2-2x)-(-x-2)(4-2x)=0

$$(1-2x)(2-2x)-(-x-2)(4-2x)=0$$

Solve for x (complex solution)

$$x=\frac{3+\sqrt{11}i}{2}\approx 1.5+1.658312395i$$

$$x=\frac{-\sqrt{11}i+3}{2}\approx 1.5-1.658312395i$$

Steps Using the Quadratic Formula

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

$$2-6x+4x^{2}-\left(-x-2\right)\left(4-2x\right)=0$$

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

$$2-6x+4x^{2}-\left(4\left(-x\right)-2\left(-x\right)x-8+4x\right)=0$$

Multiply -2 and -1 to get 2.

$$2-6x+4x^{2}-\left(4\left(-x\right)+2xx-8+4x\right)=0$$

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

$$2-6x+4x^{2}-\left(4\left(-x\right)+2x^{2}-8+4x\right)=0$$

To find the opposite of 4\left(-x\right)+2x^{2}-8+4x, find the opposite of each term.

$$2-6x+4x^{2}-4\left(-x\right)-2x^{2}+8-4x=0$$

Multiply -4 and -1 to get 4.

$$2-6x+4x^{2}+4x-2x^{2}+8-4x=0$$

Combine -6x and 4x to get -2x.

$$2-2x+4x^{2}-2x^{2}+8-4x=0$$

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

$$2-2x+2x^{2}+8-4x=0$$

Add 2 and 8 to get 10.

$$10-2x+2x^{2}-4x=0$$

Combine -2x and -4x to get -6x.

$$10-6x+2x^{2}=0$$

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.

$$2x^{2}-6x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -6 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 2\times 10}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-6\right)±\sqrt{36-8\times 10}}{2\times 2}$$

Multiply -8 times 10.

$$x=\frac{-\left(-6\right)±\sqrt{36-80}}{2\times 2}$$

Add 36 to -80.

$$x=\frac{-\left(-6\right)±\sqrt{-44}}{2\times 2}$$

Take the square root of -44.

$$x=\frac{-\left(-6\right)±2\sqrt{11}i}{2\times 2}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{11}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{6±2\sqrt{11}i}{4}$$

Now solve the equation x=\frac{6±2\sqrt{11}i}{4} when ± is plus. Add 6 to 2i\sqrt{11}.

$$x=\frac{6+2\sqrt{11}i}{4}$$

Divide 6+2i\sqrt{11} by 4.

$$x=\frac{3+\sqrt{11}i}{2}$$

Now solve the equation x=\frac{6±2\sqrt{11}i}{4} when ± is minus. Subtract 2i\sqrt{11} from 6.

$$x=\frac{-2\sqrt{11}i+6}{4}$$

Divide 6-2i\sqrt{11} by 4.

$$x=\frac{-\sqrt{11}i+3}{2}$$

The equation is now solved.

$$x=\frac{3+\sqrt{11}i}{2} x=\frac{-\sqrt{11}i+3}{2}$$

Steps for Completing the Square

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

$$2-6x+4x^{2}-\left(-x-2\right)\left(4-2x\right)=0$$

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

$$2-6x+4x^{2}-\left(4\left(-x\right)-2\left(-x\right)x-8+4x\right)=0$$

Multiply -2 and -1 to get 2.

$$2-6x+4x^{2}-\left(4\left(-x\right)+2xx-8+4x\right)=0$$

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

$$2-6x+4x^{2}-\left(4\left(-x\right)+2x^{2}-8+4x\right)=0$$

To find the opposite of 4\left(-x\right)+2x^{2}-8+4x, find the opposite of each term.

$$2-6x+4x^{2}-4\left(-x\right)-2x^{2}+8-4x=0$$

Multiply -4 and -1 to get 4.

$$2-6x+4x^{2}+4x-2x^{2}+8-4x=0$$

Combine -6x and 4x to get -2x.

$$2-2x+4x^{2}-2x^{2}+8-4x=0$$

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

$$2-2x+2x^{2}+8-4x=0$$

Add 2 and 8 to get 10.

$$10-2x+2x^{2}-4x=0$$

Combine -2x and -4x to get -6x.

$$10-6x+2x^{2}=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$-6x+2x^{2}=-10$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}-6x=-10$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -6 by 2.

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

Divide -10 by 2.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-5+\frac{9}{4}$$

Add -5 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=-\frac{11}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{-\frac{11}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{11}i}{2} x-\frac{3}{2}=-\frac{\sqrt{11}i}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{3+\sqrt{11}i}{2} x=\frac{-\sqrt{11}i+3}{2}$$