Solve x^2+20x-480=0

$${ x }^{ 2 } +20x-480=0$$

Solve for x

$$x=2\sqrt{145}-10\approx 14.083189158$$

$$x=-2\sqrt{145}-10\approx -34.083189158$$

Steps Using the Quadratic Formula

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.

$$x^{2}+20x-480=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 20 for b, and -480 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-480\right)}}{2}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-480\right)}}{2}$$

Multiply -4 times -480.

$$x=\frac{-20±\sqrt{400+1920}}{2}$$

Add 400 to 1920.

$$x=\frac{-20±\sqrt{2320}}{2}$$

Take the square root of 2320.

$$x=\frac{-20±4\sqrt{145}}{2}$$

Now solve the equation x=\frac{-20±4\sqrt{145}}{2} when ± is plus. Add -20 to 4\sqrt{145}.

$$x=\frac{4\sqrt{145}-20}{2}$$

Divide -20+4\sqrt{145} by 2.

$$x=2\sqrt{145}-10$$

Now solve the equation x=\frac{-20±4\sqrt{145}}{2} when ± is minus. Subtract 4\sqrt{145} from -20.

$$x=\frac{-4\sqrt{145}-20}{2}$$

Divide -20-4\sqrt{145} by 2.

$$x=-2\sqrt{145}-10$$

The equation is now solved.

$$x=2\sqrt{145}-10 x=-2\sqrt{145}-10$$

Steps for Completing the Square

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.

$$x^{2}+20x-480=0$$

Add 480 to both sides of the equation.

$$x^{2}+20x-480-\left(-480\right)=-\left(-480\right)$$

Subtracting -480 from itself leaves 0.

$$x^{2}+20x=-\left(-480\right)$$

Subtract -480 from 0.

$$x^{2}+20x=480$$

Divide 20, the coefficient of the x term, by 2 to get 10. Then add the square of 10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+20x+10^{2}=480+10^{2}$$

Square 10.

$$x^{2}+20x+100=480+100$$

Add 480 to 100.

$$x^{2}+20x+100=580$$

Factor x^{2}+20x+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+10\right)^{2}=580$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+10\right)^{2}}=\sqrt{580}$$

Simplify.

$$x+10=2\sqrt{145} x+10=-2\sqrt{145}$$

Subtract 10 from both sides of the equation.

$$x=2\sqrt{145}-10 x=-2\sqrt{145}-10$$