Solve x^2+20=4x+8

$${ x }^{ 2 } +20=4x+8$$

Solve for x (complex solution)

$$x=2+2\sqrt{2}i\approx 2+2.828427125i$$

$$x=-2\sqrt{2}i+2\approx 2-2.828427125i$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

$$x^{2}+20-4x=8$$

Subtract 8 from both sides.

$$x^{2}+20-4x-8=0$$

Subtract 8 from 20 to get 12.

$$x^{2}+12-4x=0$$

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.

$$x^{2}-4x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 12}}{2}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 12}}{2}$$

Multiply -4 times 12.

$$x=\frac{-\left(-4\right)±\sqrt{16-48}}{2}$$

Add 16 to -48.

$$x=\frac{-\left(-4\right)±\sqrt{-32}}{2}$$

Take the square root of -32.

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

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{2}i}{2}$$

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

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

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

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

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 4x from both sides.

$$x^{2}+20-4x=8$$

Subtract 20 from both sides.

$$x^{2}-4x=8-20$$

Subtract 20 from 8 to get -12.

$$x^{2}-4x=-12$$

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

$$x^{2}-4x+\left(-2\right)^{2}=-12+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-12+4$$

Add -12 to 4.

$$x^{2}-4x+4=-8$$

Factor x^{2}-4x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 2 to both sides of the equation.

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