Solve (0-2)x^2-2left(0+2right)x+2left(0-1right)=0

$$(0-2) { x }^{ 2 } -2 \left( 0+2 \right) x+2 \left( 0-1 \right) = 0$$

Solve for x

$$x=-1$$

Steps Using Factoring By Grouping

Subtract 2 from 0 to get -2.

$$-2x^{2}-2\left(0+2\right)x+2\left(0-1\right)=0$$

Add 0 and 2 to get 2.

$$-2x^{2}-2\times 2x+2\left(0-1\right)=0$$

Multiply 2 and 2 to get 4.

$$-2x^{2}-4x+2\left(0-1\right)=0$$

Subtract 1 from 0 to get -1.

$$-2x^{2}-4x+2\left(-1\right)=0$$

Multiply 2 and -1 to get -2.

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

Divide both sides by 2.

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

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

$$a+b=-2 ab=-\left(-1\right)=1$$

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.

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

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

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

Factor out x in -x^{2}-x.

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

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

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

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

$$x=-1 x=-1$$

Steps Using the Quadratic Formula

Subtract 2 from 0 to get -2.

$$-2x^{2}-2\left(0+2\right)x+2\left(0-1\right)=0$$

Add 0 and 2 to get 2.

$$-2x^{2}-2\times 2x+2\left(0-1\right)=0$$

Multiply 2 and 2 to get 4.

$$-2x^{2}-4x+2\left(0-1\right)=0$$

Subtract 1 from 0 to get -1.

$$-2x^{2}-4x+2\left(-1\right)=0$$

Multiply 2 and -1 to get -2.

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

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

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

Square -4.

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

Multiply -4 times -2.

$$x=\frac{-\left(-4\right)±\sqrt{16+8\left(-2\right)}}{2\left(-2\right)}$$

Multiply 8 times -2.

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

Add 16 to -16.

$$x=\frac{-\left(-4\right)±\sqrt{0}}{2\left(-2\right)}$$

Take the square root of 0.

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

The opposite of -4 is 4.

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

Multiply 2 times -2.

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

Divide 4 by -4.

$$x=-1$$

Steps for Completing the Square

Subtract 2 from 0 to get -2.

$$-2x^{2}-2\left(0+2\right)x+2\left(0-1\right)=0$$

Add 0 and 2 to get 2.

$$-2x^{2}-2\times 2x+2\left(0-1\right)=0$$

Multiply 2 and 2 to get 4.

$$-2x^{2}-4x+2\left(0-1\right)=0$$

Subtract 1 from 0 to get -1.

$$-2x^{2}-4x+2\left(-1\right)=0$$

Multiply 2 and -1 to get -2.

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

Add 2 to both sides. Anything plus zero gives itself.

$$-2x^{2}-4x=2$$

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -4 by -2.

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

Divide 2 by -2.

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

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square 1.

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

Add -1 to 1.

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

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+1=0 x+1=0$$

Subtract 1 from both sides of the equation.

$$x=-1 x=-1$$

The equation is now solved. Solutions are the same.

$$x=-1$$