Solve x^2=8x-40

$$x ^ { 2 } = 8 x - 40$$

Solve for x (complex solution)

$$x=4+2\sqrt{6}i\approx 4+4.898979486i$$

$$x=-2\sqrt{6}i+4\approx 4-4.898979486i$$

Steps Using the Quadratic Formula

Subtract 8x from both sides.

$$x^{2}-8x=-40$$

Add 40 to both sides.

$$x^{2}-8x+40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 40}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 40}}{2}$$

Multiply -4 times 40.

$$x=\frac{-\left(-8\right)±\sqrt{64-160}}{2}$$

Add 64 to -160.

$$x=\frac{-\left(-8\right)±\sqrt{-96}}{2}$$

Take the square root of -96.

$$x=\frac{-\left(-8\right)±4\sqrt{6}i}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±4\sqrt{6}i}{2}$$

Now solve the equation x=\frac{8±4\sqrt{6}i}{2} when ± is plus. Add 8 to 4i\sqrt{6}.

$$x=\frac{8+4\sqrt{6}i}{2}$$

Divide 8+4i\sqrt{6} by 2.

$$x=4+2\sqrt{6}i$$

Now solve the equation x=\frac{8±4\sqrt{6}i}{2} when ± is minus. Subtract 4i\sqrt{6} from 8.

$$x=\frac{-4\sqrt{6}i+8}{2}$$

Divide 8-4i\sqrt{6} by 2.

$$x=-2\sqrt{6}i+4$$

The equation is now solved.

$$x=4+2\sqrt{6}i x=-2\sqrt{6}i+4$$

Steps for Completing the Square

Subtract 8x from both sides.

$$x^{2}-8x=-40$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-40+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-40+16$$

Add -40 to 16.

$$x^{2}-8x+16=-24$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=-24$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{-24}$$

Simplify.

$$x-4=2\sqrt{6}i x-4=-2\sqrt{6}i$$

Add 4 to both sides of the equation.

$$x=4+2\sqrt{6}i x=-2\sqrt{6}i+4$$