Solve 3x^2+13x+12=0

$$3 { x }^{ 2 } +13x+12 = 0$$

Solve for x

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

$$x=-3$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=13 ab=3\times 12=36$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 36.

$$1,36 2,18 3,12 4,9 6,6$$

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 13.

$$a=4 b=9$$

Rewrite 3x^{2}+13x+12 as \left(3x^{2}+4x\right)+\left(9x+12\right).

$$\left(3x^{2}+4x\right)+\left(9x+12\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(3x+4\right)+3\left(3x+4\right)$$

Factor out common term 3x+4 by using distributive property.

$$\left(3x+4\right)\left(x+3\right)$$

To find equation solutions, solve 3x+4=0 and x+3=0.

$$x=-\frac{4}{3} x=-3$$

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.

$$3x^{2}+13x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 13 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-13±\sqrt{13^{2}-4\times 3\times 12}}{2\times 3}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\times 3\times 12}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-13±\sqrt{169-12\times 12}}{2\times 3}$$

Multiply -12 times 12.

$$x=\frac{-13±\sqrt{169-144}}{2\times 3}$$

Add 169 to -144.

$$x=\frac{-13±\sqrt{25}}{2\times 3}$$

Take the square root of 25.

$$x=\frac{-13±5}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-13±5}{6}$$

Now solve the equation x=\frac{-13±5}{6} when ± is plus. Add -13 to 5.

$$x=-\frac{8}{6}$$

Reduce the fraction \frac{-8}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{4}{3}$$

Now solve the equation x=\frac{-13±5}{6} when ± is minus. Subtract 5 from -13.

$$x=-\frac{18}{6}$$

Divide -18 by 6.

$$x=-3$$

The equation is now solved.

$$x=-\frac{4}{3} x=-3$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$3x^{2}+13x+12=0$$

Subtract 12 from both sides of the equation.

$$3x^{2}+13x+12-12=-12$$

Subtracting 12 from itself leaves 0.

$$3x^{2}+13x=-12$$

Divide both sides by 3.

$$\frac{3x^{2}+13x}{3}=-\frac{12}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{13}{3}x=-\frac{12}{3}$$

Divide -12 by 3.

$$x^{2}+\frac{13}{3}x=-4$$

Divide \frac{13}{3}, the coefficient of the x term, by 2 to get \frac{13}{6}. Then add the square of \frac{13}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{13}{3}x+\left(\frac{13}{6}\right)^{2}=-4+\left(\frac{13}{6}\right)^{2}$$

Square \frac{13}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{13}{3}x+\frac{169}{36}=-4+\frac{169}{36}$$

Add -4 to \frac{169}{36}.

$$x^{2}+\frac{13}{3}x+\frac{169}{36}=\frac{25}{36}$$

Factor x^{2}+\frac{13}{3}x+\frac{169}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{13}{6}\right)^{2}=\frac{25}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Simplify.

$$x+\frac{13}{6}=\frac{5}{6} x+\frac{13}{6}=-\frac{5}{6}$$

Subtract \frac{13}{6} from both sides of the equation.

$$x=-\frac{4}{3} x=-3$$