Solve 16+8x=8x^2

$$16 + 8 x = 8 x ^ { 2 }$$

Solve for x

$$x=-1$$

$$x=2$$

Steps Using Factoring By Grouping

Subtract 8x^{2} from both sides.

$$16+8x-8x^{2}=0$$

Divide both sides by 8.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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

$$a+b=1 ab=-2=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=2 b=-1$$

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

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

Factor out -x in the first and -1 in the second group.

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

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

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

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

$$x=2 x=-1$$

Steps Using the Quadratic Formula

Subtract 8x^{2} from both sides.

$$16+8x-8x^{2}=0$$

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.

$$-8x^{2}+8x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 8 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-8\right)\times 16}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-8±\sqrt{64+32\times 16}}{2\left(-8\right)}$$

Multiply 32 times 16.

$$x=\frac{-8±\sqrt{64+512}}{2\left(-8\right)}$$

Add 64 to 512.

$$x=\frac{-8±\sqrt{576}}{2\left(-8\right)}$$

Take the square root of 576.

$$x=\frac{-8±24}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{-8±24}{-16}$$

Now solve the equation x=\frac{-8±24}{-16} when ± is plus. Add -8 to 24.

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

Divide 16 by -16.

$$x=-1$$

Now solve the equation x=\frac{-8±24}{-16} when ± is minus. Subtract 24 from -8.

$$x=-\frac{32}{-16}$$

Divide -32 by -16.

$$x=2$$

The equation is now solved.

$$x=-1 x=2$$

Steps for Completing the Square

Subtract 8x^{2} from both sides.

$$16+8x-8x^{2}=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by -8.

$$\frac{-8x^{2}+8x}{-8}=-\frac{16}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x^{2}+\frac{8}{-8}x=-\frac{16}{-8}$$

Divide 8 by -8.

$$x^{2}-x=-\frac{16}{-8}$$

Divide -16 by -8.

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

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

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=2+\frac{1}{4}$$

Add 2 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{9}{4}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{3}{2} x-\frac{1}{2}=-\frac{3}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=2 x=-1$$