Solve 0=3left(x+3right)^2-12

$$0=3 { \left(x+3 \right) }^{ 2 } -12$$

Solve for x

$$x=-1$$

$$x=-5$$

Steps Using Factoring By Grouping

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

$$0=3\left(x^{2}+6x+9\right)-12$$

Use the distributive property to multiply 3 by x^{2}+6x+9.

$$0=3x^{2}+18x+27-12$$

Subtract 12 from 27 to get 15.

$$0=3x^{2}+18x+15$$

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

$$3x^{2}+18x+15=0$$

Divide both sides by 3.

$$x^{2}+6x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\times 5=5$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=5$$

Rewrite x^{2}+6x+5 as \left(x^{2}+x\right)+\left(5x+5\right).

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

Factor out x in the first and 5 in the second group.

$$x\left(x+1\right)+5\left(x+1\right)$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(x+5\right)$$

To find equation solutions, solve x+1=0 and x+5=0.

$$x=-1 x=-5$$

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}.

$$0=3\left(x^{2}+6x+9\right)-12$$

Use the distributive property to multiply 3 by x^{2}+6x+9.

$$0=3x^{2}+18x+27-12$$

Subtract 12 from 27 to get 15.

$$0=3x^{2}+18x+15$$

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

$$3x^{2}+18x+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 18 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\times 3\times 15}}{2\times 3}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\times 3\times 15}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-18±\sqrt{324-12\times 15}}{2\times 3}$$

Multiply -12 times 15.

$$x=\frac{-18±\sqrt{324-180}}{2\times 3}$$

Add 324 to -180.

$$x=\frac{-18±\sqrt{144}}{2\times 3}$$

Take the square root of 144.

$$x=\frac{-18±12}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-18±12}{6}$$

Now solve the equation x=\frac{-18±12}{6} when ± is plus. Add -18 to 12.

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

Divide -6 by 6.

$$x=-1$$

Now solve the equation x=\frac{-18±12}{6} when ± is minus. Subtract 12 from -18.

$$x=-\frac{30}{6}$$

Divide -30 by 6.

$$x=-5$$

The equation is now solved.

$$x=-1 x=-5$$

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}.

$$0=3\left(x^{2}+6x+9\right)-12$$

Use the distributive property to multiply 3 by x^{2}+6x+9.

$$0=3x^{2}+18x+27-12$$

Subtract 12 from 27 to get 15.

$$0=3x^{2}+18x+15$$

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

$$3x^{2}+18x+15=0$$

Subtract 15 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}+18x=-15$$

Divide both sides by 3.

$$\frac{3x^{2}+18x}{3}=-\frac{15}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{18}{3}x=-\frac{15}{3}$$

Divide 18 by 3.

$$x^{2}+6x=-\frac{15}{3}$$

Divide -15 by 3.

$$x^{2}+6x=-5$$

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.

$$x^{2}+6x+3^{2}=-5+3^{2}$$

Square 3.

$$x^{2}+6x+9=-5+9$$

Add -5 to 9.

$$x^{2}+6x+9=4$$

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}.

$$\left(x+3\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+3=2 x+3=-2$$

Subtract 3 from both sides of the equation.

$$x=-1 x=-5$$