Solve x^2+6x+9=12

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

Solve for x

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

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

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

Subtract 12 from both sides of the equation.

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

Subtracting 12 from itself leaves 0.

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

Subtract 12 from 9.

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

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}.

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

Square 6.

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

Multiply -4 times -3.

$$x=\frac{-6±\sqrt{36+12}}{2}$$

Add 36 to 12.

$$x=\frac{-6±\sqrt{48}}{2}$$

Take the square root of 48.

$$x=\frac{-6±4\sqrt{3}}{2}$$

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.

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

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

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

The equation is now solved.

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

Steps for Completing the Square

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.

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

Subtract 3 from both sides of the equation.

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