Solve x^2=4x+4^2+20^2

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

Solve for x

$$x=2\sqrt{105}+2\approx 22.493901532$$

$$x=2-2\sqrt{105}\approx -18.493901532$$

Steps Using the Quadratic Formula

Calculate 4 to the power of 2 and get 16.

$$x^{2}=4x+16+20^{2}$$

Calculate 20 to the power of 2 and get 400.

$$x^{2}=4x+16+400$$

Add 16 and 400 to get 416.

$$x^{2}=4x+416$$

Subtract 4x from both sides.

$$x^{2}-4x=416$$

Subtract 416 from both sides.

$$x^{2}-4x-416=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and -416 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-416\right)}}{2}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-416\right)}}{2}$$

Multiply -4 times -416.

$$x=\frac{-\left(-4\right)±\sqrt{16+1664}}{2}$$

Add 16 to 1664.

$$x=\frac{-\left(-4\right)±\sqrt{1680}}{2}$$

Take the square root of 1680.

$$x=\frac{-\left(-4\right)±4\sqrt{105}}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{105}}{2}$$

Now solve the equation x=\frac{4±4\sqrt{105}}{2} when ± is plus. Add 4 to 4\sqrt{105}.

$$x=\frac{4\sqrt{105}+4}{2}$$

Divide 4+4\sqrt{105} by 2.

$$x=2\sqrt{105}+2$$

Now solve the equation x=\frac{4±4\sqrt{105}}{2} when ± is minus. Subtract 4\sqrt{105} from 4.

$$x=\frac{4-4\sqrt{105}}{2}$$

Divide 4-4\sqrt{105} by 2.

$$x=2-2\sqrt{105}$$

The equation is now solved.

$$x=2\sqrt{105}+2 x=2-2\sqrt{105}$$

Steps for Completing the Square

Calculate 4 to the power of 2 and get 16.

$$x^{2}=4x+16+20^{2}$$

Calculate 20 to the power of 2 and get 400.

$$x^{2}=4x+16+400$$

Add 16 and 400 to get 416.

$$x^{2}=4x+416$$

Subtract 4x from both sides.

$$x^{2}-4x=416$$

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.

$$x^{2}-4x+\left(-2\right)^{2}=416+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=416+4$$

Add 416 to 4.

$$x^{2}-4x+4=420$$

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}.

$$\left(x-2\right)^{2}=420$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{420}$$

Simplify.

$$x-2=2\sqrt{105} x-2=-2\sqrt{105}$$

Add 2 to both sides of the equation.

$$x=2\sqrt{105}+2 x=2-2\sqrt{105}$$