Solve x^2-16x+56=0

$${ x }^{ 2 } -16x+56 = 0$$

Solve for x

$$x=2\sqrt{2}+8\approx 10.828427125$$

$$x=8-2\sqrt{2}\approx 5.171572875$$

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.

$$x^{2}-16x+56=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -16 for b, and 56 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 56}}{2}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 56}}{2}$$

Multiply -4 times 56.

$$x=\frac{-\left(-16\right)±\sqrt{256-224}}{2}$$

Add 256 to -224.

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

Take the square root of 32.

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

The opposite of -16 is 16.

$$x=\frac{16±4\sqrt{2}}{2}$$

Now solve the equation x=\frac{16±4\sqrt{2}}{2} when ± is plus. Add 16 to 4\sqrt{2}.

$$x=\frac{4\sqrt{2}+16}{2}$$

Divide 16+4\sqrt{2} by 2.

$$x=2\sqrt{2}+8$$

Now solve the equation x=\frac{16±4\sqrt{2}}{2} when ± is minus. Subtract 4\sqrt{2} from 16.

$$x=\frac{16-4\sqrt{2}}{2}$$

Divide 16-4\sqrt{2} by 2.

$$x=8-2\sqrt{2}$$

The equation is now solved.

$$x=2\sqrt{2}+8 x=8-2\sqrt{2}$$

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.

$$x^{2}-16x+56=0$$

Subtract 56 from both sides of the equation.

$$x^{2}-16x+56-56=-56$$

Subtracting 56 from itself leaves 0.

$$x^{2}-16x=-56$$

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.

$$x^{2}-16x+\left(-8\right)^{2}=-56+\left(-8\right)^{2}$$

Square -8.

$$x^{2}-16x+64=-56+64$$

Add -56 to 64.

$$x^{2}-16x+64=8$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-8=2\sqrt{2} x-8=-2\sqrt{2}$$

Add 8 to both sides of the equation.

$$x=2\sqrt{2}+8 x=8-2\sqrt{2}$$