Solve 0=-16x^2+16x+480

$$0 = -16 { x }^{ 2 } +16x+480$$

Solve for x

$$x=-5$$

$$x=6$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$-16x^{2}+16x+480=0$$

Divide both sides by 16.

$$-x^{2}+x+30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+30. To find a and b, set up a system to be solved.

$$a+b=1 ab=-30=-30$$

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. List all such integer pairs that give product -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=6 b=-5$$

Rewrite -x^{2}+x+30 as \left(-x^{2}+6x\right)+\left(-5x+30\right).

$$\left(-x^{2}+6x\right)+\left(-5x+30\right)$$

Factor out -x in the first and -5 in the second group.

$$-x\left(x-6\right)-5\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(-x-5\right)$$

To find equation solutions, solve x-6=0 and -x-5=0.

$$x=6 x=-5$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$-16x^{2}+16x+480=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 16 for b, and 480 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\left(-16\right)\times 480}}{2\left(-16\right)}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\left(-16\right)\times 480}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$x=\frac{-16±\sqrt{256+64\times 480}}{2\left(-16\right)}$$

Multiply 64 times 480.

$$x=\frac{-16±\sqrt{256+30720}}{2\left(-16\right)}$$

Add 256 to 30720.

$$x=\frac{-16±\sqrt{30976}}{2\left(-16\right)}$$

Take the square root of 30976.

$$x=\frac{-16±176}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{-16±176}{-32}$$

Now solve the equation x=\frac{-16±176}{-32} when ± is plus. Add -16 to 176.

$$x=\frac{160}{-32}$$

Divide 160 by -32.

$$x=-5$$

Now solve the equation x=\frac{-16±176}{-32} when ± is minus. Subtract 176 from -16.

$$x=-\frac{192}{-32}$$

Divide -192 by -32.

$$x=6$$

The equation is now solved.

$$x=-5 x=6$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$-16x^{2}+16x+480=0$$

Subtract 480 from both sides. Anything subtracted from zero gives its negation.

$$-16x^{2}+16x=-480$$

Divide both sides by -16.

$$\frac{-16x^{2}+16x}{-16}=-\frac{480}{-16}$$

Dividing by -16 undoes the multiplication by -16.

$$x^{2}+\frac{16}{-16}x=-\frac{480}{-16}$$

Divide 16 by -16.

$$x^{2}-x=-\frac{480}{-16}$$

Divide -480 by -16.

$$x^{2}-x=30$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=30+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=30+\frac{1}{4}$$

Add 30 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{121}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{11}{2} x-\frac{1}{2}=-\frac{11}{2}$$

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

$$x=6 x=-5$$