Solve x^2=2x+4

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

Solve for x

$$x=\sqrt{5}+1\approx 3.236067977$$

$$x=1-\sqrt{5}\approx -1.236067977$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

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

Subtract 4 from both sides.

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

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

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

Square -2.

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

Multiply -4 times -4.

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

Add 4 to 16.

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

Take the square root of 20.

$$x=\frac{-\left(-2\right)±2\sqrt{5}}{2}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{5}}{2}$$

Now solve the equation x=\frac{2±2\sqrt{5}}{2} when ± is plus. Add 2 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}+2}{2}$$

Divide 2+2\sqrt{5} by 2.

$$x=\sqrt{5}+1$$

Now solve the equation x=\frac{2±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from 2.

$$x=\frac{2-2\sqrt{5}}{2}$$

Divide 2-2\sqrt{5} by 2.

$$x=1-\sqrt{5}$$

The equation is now solved.

$$x=\sqrt{5}+1 x=1-\sqrt{5}$$

Steps for Completing the Square

Subtract 2x from both sides.

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

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=4+1$$

Add 4 to 1.

$$x^{2}-2x+1=5$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{5}$$

Simplify.

$$x-1=\sqrt{5} x-1=-\sqrt{5}$$

Add 1 to both sides of the equation.

$$x=\sqrt{5}+1 x=1-\sqrt{5}$$