Solve 5(x-5)^2=125

$$5 ( x - 5 ) ^ { 2 } = 125$$

Solve for x

$$x=10$$

$$x=0$$

Steps Using Factoring

Divide both sides by 5.

$$\left(x-5\right)^{2}=\frac{125}{5}$$

Divide 125 by 5 to get 25.

$$\left(x-5\right)^{2}=25$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-10x+25=25$$

Subtract 25 from both sides.

$$x^{2}-10x+25-25=0$$

Subtract 25 from 25 to get 0.

$$x^{2}-10x=0$$

Factor out x.

$$x\left(x-10\right)=0$$

To find equation solutions, solve x=0 and x-10=0.

$$x=0 x=10$$

Steps Using the Quadratic Formula

Divide both sides by 5.

$$\left(x-5\right)^{2}=\frac{125}{5}$$

Divide 125 by 5 to get 25.

$$\left(x-5\right)^{2}=25$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-10x+25=25$$

Subtract 25 from both sides.

$$x^{2}-10x+25-25=0$$

Subtract 25 from 25 to get 0.

$$x^{2}-10x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}}}{2}$$

Take the square root of \left(-10\right)^{2}.

$$x=\frac{-\left(-10\right)±10}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±10}{2}$$

Now solve the equation x=\frac{10±10}{2} when ± is plus. Add 10 to 10.

$$x=\frac{20}{2}$$

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{10±10}{2} when ± is minus. Subtract 10 from 10.

$$x=\frac{0}{2}$$

Divide 0 by 2.

$$x=0$$

The equation is now solved.

$$x=10 x=0$$

Steps for Completing the Square

Divide both sides by 5.

$$\left(x-5\right)^{2}=\frac{125}{5}$$

Divide 125 by 5 to get 25.

$$\left(x-5\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{25}$$

Simplify.

$$x-5=5 x-5=-5$$

Add 5 to both sides of the equation.

$$x=10 x=0$$