Solve x^2-16x-128=0

$${ x }^{ 2 } -16x-128=0$$

Solve for x

$$x=8\sqrt{3}+8\approx 21.856406461$$

$$x=8-8\sqrt{3}\approx -5.856406461$$

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.

$$x^{2}-16x-128=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -16 for b, and -128 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\left(-128\right)}}{2}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\left(-128\right)}}{2}$$

Multiply -4 times -128.

$$x=\frac{-\left(-16\right)±\sqrt{256+512}}{2}$$

Add 256 to 512.

$$x=\frac{-\left(-16\right)±\sqrt{768}}{2}$$

Take the square root of 768.

$$x=\frac{-\left(-16\right)±16\sqrt{3}}{2}$$

The opposite of -16 is 16.

$$x=\frac{16±16\sqrt{3}}{2}$$

Now solve the equation x=\frac{16±16\sqrt{3}}{2} when ± is plus. Add 16 to 16\sqrt{3}.

$$x=\frac{16\sqrt{3}+16}{2}$$

Divide 16+16\sqrt{3} by 2.

$$x=8\sqrt{3}+8$$

Now solve the equation x=\frac{16±16\sqrt{3}}{2} when ± is minus. Subtract 16\sqrt{3} from 16.

$$x=\frac{16-16\sqrt{3}}{2}$$

Divide 16-16\sqrt{3} by 2.

$$x=8-8\sqrt{3}$$

The equation is now solved.

$$x=8\sqrt{3}+8 x=8-8\sqrt{3}$$

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.

$$x^{2}-16x-128=0$$

Add 128 to both sides of the equation.

$$x^{2}-16x-128-\left(-128\right)=-\left(-128\right)$$

Subtracting -128 from itself leaves 0.

$$x^{2}-16x=-\left(-128\right)$$

Subtract -128 from 0.

$$x^{2}-16x=128$$

Divide -16, the coefficient of the x term, by 2 to get -8. Then add the square of -8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-16x+\left(-8\right)^{2}=128+\left(-8\right)^{2}$$

Square -8.

$$x^{2}-16x+64=128+64$$

Add 128 to 64.

$$x^{2}-16x+64=192$$

Factor x^{2}-16x+64. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-8\right)^{2}=192$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-8\right)^{2}}=\sqrt{192}$$

Simplify.

$$x-8=8\sqrt{3} x-8=-8\sqrt{3}$$

Add 8 to both sides of the equation.

$$x=8\sqrt{3}+8 x=8-8\sqrt{3}$$