Solve 2x^2-8x=4x-16

$$2 x ^ { 2 } - 8 x = 4 x - 16$$

Solve for x

$$x=2$$

$$x=4$$

Steps Using Factoring By Grouping

Subtract 4x from both sides.

$$2x^{2}-8x-4x=-16$$

Combine -8x and -4x to get -12x.

$$2x^{2}-12x=-16$$

Add 16 to both sides.

$$2x^{2}-12x+16=0$$

Divide both sides by 2.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\times 8=8$$

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 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -6.

$$a=-4 b=-2$$

Rewrite x^{2}-6x+8 as \left(x^{2}-4x\right)+\left(-2x+8\right).

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

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

$$x\left(x-4\right)-2\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x-2\right)$$

To find equation solutions, solve x-4=0 and x-2=0.

$$x=4 x=2$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

$$2x^{2}-8x-4x=-16$$

Combine -8x and -4x to get -12x.

$$2x^{2}-12x=-16$$

Add 16 to both sides.

$$2x^{2}-12x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -12 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\times 16}}{2\times 2}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 2\times 16}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-12\right)±\sqrt{144-8\times 16}}{2\times 2}$$

Multiply -8 times 16.

$$x=\frac{-\left(-12\right)±\sqrt{144-128}}{2\times 2}$$

Add 144 to -128.

$$x=\frac{-\left(-12\right)±\sqrt{16}}{2\times 2}$$

Take the square root of 16.

$$x=\frac{-\left(-12\right)±4}{2\times 2}$$

The opposite of -12 is 12.

$$x=\frac{12±4}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{12±4}{4}$$

Now solve the equation x=\frac{12±4}{4} when ± is plus. Add 12 to 4.

$$x=\frac{16}{4}$$

Divide 16 by 4.

$$x=4$$

Now solve the equation x=\frac{12±4}{4} when ± is minus. Subtract 4 from 12.

$$x=\frac{8}{4}$$

Divide 8 by 4.

$$x=2$$

The equation is now solved.

$$x=4 x=2$$

Steps for Completing the Square

Subtract 4x from both sides.

$$2x^{2}-8x-4x=-16$$

Combine -8x and -4x to get -12x.

$$2x^{2}-12x=-16$$

Divide both sides by 2.

$$\frac{2x^{2}-12x}{2}=-\frac{16}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{12}{2}\right)x=-\frac{16}{2}$$

Divide -12 by 2.

$$x^{2}-6x=-\frac{16}{2}$$

Divide -16 by 2.

$$x^{2}-6x=-8$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-8+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-8+9$$

Add -8 to 9.

$$x^{2}-6x+9=1$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-3=1 x-3=-1$$

Add 3 to both sides of the equation.

$$x=4 x=2$$