Solve 3x^2+9x+6=0quadx

$$3 x ^ { 2 } + 9 x + 6 = 0 \quad x$$

Solve for x

$$x=-2$$

$$x=-1$$

Steps Using Factoring By Grouping

Anything times zero gives zero.

$$3x^{2}+9x+6=0$$

Divide both sides by 3.

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

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

$$a+b=3 ab=1\times 2=2$$

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.

$$a=1 b=2$$

Rewrite x^{2}+3x+2 as \left(x^{2}+x\right)+\left(2x+2\right).

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

Factor out x in the first and 2 in the second group.

$$x\left(x+1\right)+2\left(x+1\right)$$

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

$$\left(x+1\right)\left(x+2\right)$$

To find equation solutions, solve x+1=0 and x+2=0.

$$x=-1 x=-2$$

Steps Using the Quadratic Formula

Anything times zero gives zero.

$$3x^{2}+9x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 9 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\times 3\times 6}}{2\times 3}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 3\times 6}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-9±\sqrt{81-12\times 6}}{2\times 3}$$

Multiply -12 times 6.

$$x=\frac{-9±\sqrt{81-72}}{2\times 3}$$

Add 81 to -72.

$$x=\frac{-9±\sqrt{9}}{2\times 3}$$

Take the square root of 9.

$$x=\frac{-9±3}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-9±3}{6}$$

Now solve the equation x=\frac{-9±3}{6} when ± is plus. Add -9 to 3.

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

Divide -6 by 6.

$$x=-1$$

Now solve the equation x=\frac{-9±3}{6} when ± is minus. Subtract 3 from -9.

$$x=-\frac{12}{6}$$

Divide -12 by 6.

$$x=-2$$

The equation is now solved.

$$x=-1 x=-2$$

Steps for Completing the Square

Anything times zero gives zero.

$$3x^{2}+9x+6=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide 9 by 3.

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

Divide -6 by 3.

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

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=-2+\frac{9}{4}$$

Add -2 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{1}{4}$$

Factor x^{2}+3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{1}{2} x+\frac{3}{2}=-\frac{1}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=-1 x=-2$$