Solve 3(x-5)^2-75=0

$$3 ( x - 5 ) ^ { 2 } - 75 = 0$$

Solve for x

$$x=10$$

$$x=0$$

Steps Using Factoring

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

$$3\left(x^{2}-10x+25\right)-75=0$$

Use the distributive property to multiply 3 by x^{2}-10x+25.

$$3x^{2}-30x+75-75=0$$

Subtract 75 from 75 to get 0.

$$3x^{2}-30x=0$$

Factor out x.

$$x\left(3x-30\right)=0$$

To find equation solutions, solve x=0 and 3x-30=0.

$$x=0 x=10$$

Steps Using the Quadratic Formula

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

$$3\left(x^{2}-10x+25\right)-75=0$$

Use the distributive property to multiply 3 by x^{2}-10x+25.

$$3x^{2}-30x+75-75=0$$

Subtract 75 from 75 to get 0.

$$3x^{2}-30x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -30 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}}}{2\times 3}$$

Take the square root of \left(-30\right)^{2}.

$$x=\frac{-\left(-30\right)±30}{2\times 3}$$

The opposite of -30 is 30.

$$x=\frac{30±30}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{30±30}{6}$$

Now solve the equation x=\frac{30±30}{6} when ± is plus. Add 30 to 30.

$$x=\frac{60}{6}$$

Divide 60 by 6.

$$x=10$$

Now solve the equation x=\frac{30±30}{6} when ± is minus. Subtract 30 from 30.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

The equation is now solved.

$$x=10 x=0$$

Steps for Completing the Square

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

$$3\left(x^{2}-10x+25\right)-75=0$$

Use the distributive property to multiply 3 by x^{2}-10x+25.

$$3x^{2}-30x+75-75=0$$

Subtract 75 from 75 to get 0.

$$3x^{2}-30x=0$$

Divide both sides by 3.

$$\frac{3x^{2}-30x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{30}{3}\right)x=\frac{0}{3}$$

Divide -30 by 3.

$$x^{2}-10x=\frac{0}{3}$$

Divide 0 by 3.

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

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=\left(-5\right)^{2}$$

Square -5.

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

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

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.

$$x=10 x=0$$