Solve x^2-20x=-2x-80

$$x ^ { 2 } - 20 x = - 2 x - 80$$

Solve for x

$$x=8$$

$$x=10$$

Steps Using Factoring

Add 2x to both sides.

$$x^{2}-20x+2x=-80$$

Combine -20x and 2x to get -18x.

$$x^{2}-18x=-80$$

Add 80 to both sides.

$$x^{2}-18x+80=0$$

To solve the equation, factor x^{2}-18x+80 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-18 ab=80$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 80.

$$-1,-80 -2,-40 -4,-20 -5,-16 -8,-10$$

Calculate the sum for each pair.

$$-1-80=-81 -2-40=-42 -4-20=-24 -5-16=-21 -8-10=-18$$

The solution is the pair that gives sum -18.

$$a=-10 b=-8$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-10\right)\left(x-8\right)$$

To find equation solutions, solve x-10=0 and x-8=0.

$$x=10 x=8$$

Steps Using Factoring By Grouping

Add 2x to both sides.

$$x^{2}-20x+2x=-80$$

Combine -20x and 2x to get -18x.

$$x^{2}-18x=-80$$

Add 80 to both sides.

$$x^{2}-18x+80=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+80. To find a and b, set up a system to be solved.

$$a+b=-18 ab=1\times 80=80$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 80.

$$-1,-80 -2,-40 -4,-20 -5,-16 -8,-10$$

Calculate the sum for each pair.

$$-1-80=-81 -2-40=-42 -4-20=-24 -5-16=-21 -8-10=-18$$

The solution is the pair that gives sum -18.

$$a=-10 b=-8$$

Rewrite x^{2}-18x+80 as \left(x^{2}-10x\right)+\left(-8x+80\right).

$$\left(x^{2}-10x\right)+\left(-8x+80\right)$$

Factor out x in the first and -8 in the second group.

$$x\left(x-10\right)-8\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x-8\right)$$

To find equation solutions, solve x-10=0 and x-8=0.

$$x=10 x=8$$

Steps Using the Quadratic Formula

Add 2x to both sides.

$$x^{2}-20x+2x=-80$$

Combine -20x and 2x to get -18x.

$$x^{2}-18x=-80$$

Add 80 to both sides.

$$x^{2}-18x+80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -18 for b, and 80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 80}}{2}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 80}}{2}$$

Multiply -4 times 80.

$$x=\frac{-\left(-18\right)±\sqrt{324-320}}{2}$$

Add 324 to -320.

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

Take the square root of 4.

$$x=\frac{-\left(-18\right)±2}{2}$$

The opposite of -18 is 18.

$$x=\frac{18±2}{2}$$

Now solve the equation x=\frac{18±2}{2} when ± is plus. Add 18 to 2.

$$x=\frac{20}{2}$$

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{18±2}{2} when ± is minus. Subtract 2 from 18.

$$x=\frac{16}{2}$$

Divide 16 by 2.

$$x=8$$

The equation is now solved.

$$x=10 x=8$$

Steps for Completing the Square

Add 2x to both sides.

$$x^{2}-20x+2x=-80$$

Combine -20x and 2x to get -18x.

$$x^{2}-18x=-80$$

Divide -18, the coefficient of the x term, by 2 to get -9. Then add the square of -9 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-18x+\left(-9\right)^{2}=-80+\left(-9\right)^{2}$$

Square -9.

$$x^{2}-18x+81=-80+81$$

Add -80 to 81.

$$x^{2}-18x+81=1$$

Factor x^{2}-18x+81. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-9\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-9\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-9=1 x-9=-1$$

Add 9 to both sides of the equation.

$$x=10 x=8$$