Solve 20^2=x(x-6)

$${ 20 }^{ 2 } =x(x-6)$$

Solve for x

$$x=\sqrt{409}+3\approx 23.223748416$$

$$x=3-\sqrt{409}\approx -17.223748416$$

Steps Using the Quadratic Formula

Calculate 20 to the power of 2 and get 400.

$$400=x\left(x-6\right)$$

Use the distributive property to multiply x by x-6.

$$400=x^{2}-6x$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-6x=400$$

Subtract 400 from both sides.

$$x^{2}-6x-400=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and -400 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-400\right)}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-400\right)}}{2}$$

Multiply -4 times -400.

$$x=\frac{-\left(-6\right)±\sqrt{36+1600}}{2}$$

Add 36 to 1600.

$$x=\frac{-\left(-6\right)±\sqrt{1636}}{2}$$

Take the square root of 1636.

$$x=\frac{-\left(-6\right)±2\sqrt{409}}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{409}}{2}$$

Now solve the equation x=\frac{6±2\sqrt{409}}{2} when ± is plus. Add 6 to 2\sqrt{409}.

$$x=\frac{2\sqrt{409}+6}{2}$$

Divide 6+2\sqrt{409} by 2.

$$x=\sqrt{409}+3$$

Now solve the equation x=\frac{6±2\sqrt{409}}{2} when ± is minus. Subtract 2\sqrt{409} from 6.

$$x=\frac{6-2\sqrt{409}}{2}$$

Divide 6-2\sqrt{409} by 2.

$$x=3-\sqrt{409}$$

The equation is now solved.

$$x=\sqrt{409}+3 x=3-\sqrt{409}$$

Steps for Completing the Square

Calculate 20 to the power of 2 and get 400.

$$400=x\left(x-6\right)$$

Use the distributive property to multiply x by x-6.

$$400=x^{2}-6x$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-6x=400$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=400+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=400+9$$

Add 400 to 9.

$$x^{2}-6x+9=409$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=409$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{409}$$

Simplify.

$$x-3=\sqrt{409} x-3=-\sqrt{409}$$

Add 3 to both sides of the equation.

$$x=\sqrt{409}+3 x=3-\sqrt{409}$$