Solve 0=(x-3)^2-12

$$0 = ( x - 3 ) ^ { 2 } - 12$$

Solve for x

$$x=2\sqrt{3}+3\approx 6.464101615$$

$$x=3-2\sqrt{3}\approx -0.464101615$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$0=x^{2}-6x+9-12$$

Subtract 12 from 9 to get -3.

$$0=x^{2}-6x-3$$

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

$$x^{2}-6x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-3\right)}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-3\right)}}{2}$$

Multiply -4 times -3.

$$x=\frac{-\left(-6\right)±\sqrt{36+12}}{2}$$

Add 36 to 12.

$$x=\frac{-\left(-6\right)±\sqrt{48}}{2}$$

Take the square root of 48.

$$x=\frac{-\left(-6\right)±4\sqrt{3}}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±4\sqrt{3}}{2}$$

Now solve the equation x=\frac{6±4\sqrt{3}}{2} when ± is plus. Add 6 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}+6}{2}$$

Divide 6+4\sqrt{3} by 2.

$$x=2\sqrt{3}+3$$

Now solve the equation x=\frac{6±4\sqrt{3}}{2} when ± is minus. Subtract 4\sqrt{3} from 6.

$$x=\frac{6-4\sqrt{3}}{2}$$

Divide 6-4\sqrt{3} by 2.

$$x=3-2\sqrt{3}$$

The equation is now solved.

$$x=2\sqrt{3}+3 x=3-2\sqrt{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$0=x^{2}-6x+9-12$$

Subtract 12 from 9 to get -3.

$$0=x^{2}-6x-3$$

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

$$x^{2}-6x-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$x^{2}-6x=3$$

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.

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

Square -3.

$$x^{2}-6x+9=3+9$$

Add 3 to 9.

$$x^{2}-6x+9=12$$

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}.

$$\left(x-3\right)^{2}=12$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{12}$$

Simplify.

$$x-3=2\sqrt{3} x-3=-2\sqrt{3}$$

Add 3 to both sides of the equation.

$$x=2\sqrt{3}+3 x=3-2\sqrt{3}$$