Solve 3x^2+2x=12-3x

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

Solve for x

$$x=-3$$

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

Steps Using Factoring By Grouping

Subtract 12 from both sides.

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

Add 3x to both sides.

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

Combine 2x and 3x to get 5x.

$$3x^{2}+5x-12=0$$

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.

$$a+b=5 ab=3\left(-12\right)=-36$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. 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.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 5.

$$a=-4 b=9$$

Rewrite 3x^{2}+5x-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

Subtract 12 from both sides.

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

Add 3x to both sides.

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

Combine 2x and 3x to get 5x.

$$3x^{2}+5x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 5 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 3\left(-12\right)}}{2\times 3}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 3\left(-12\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-5±\sqrt{25-12\left(-12\right)}}{2\times 3}$$

Multiply -12 times -12.

$$x=\frac{-5±\sqrt{25+144}}{2\times 3}$$

Add 25 to 144.

$$x=\frac{-5±\sqrt{169}}{2\times 3}$$

Take the square root of 169.

$$x=\frac{-5±13}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-5±13}{6}$$

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

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

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{-5±13}{6} when ± is minus. Subtract 13 from -5.

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

Divide -18 by 6.

$$x=-3$$

The equation is now solved.

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

Steps for Completing the Square

Add 3x to both sides.

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

Combine 2x and 3x to get 5x.

$$3x^{2}+5x=12$$

Divide both sides by 3.

$$\frac{3x^{2}+5x}{3}=\frac{12}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{5}{3}x=\frac{12}{3}$$

Divide 12 by 3.

$$x^{2}+\frac{5}{3}x=4$$

Divide \frac{5}{3}, the coefficient of the x term, by 2 to get \frac{5}{6}. Then add the square of \frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{3}x+\left(\frac{5}{6}\right)^{2}=4+\left(\frac{5}{6}\right)^{2}$$

Square \frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=4+\frac{25}{36}$$

Add 4 to \frac{25}{36}.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=\frac{169}{36}$$

Factor x^{2}+\frac{5}{3}x+\frac{25}{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}.

$$\left(x+\frac{5}{6}\right)^{2}=\frac{169}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{6}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Simplify.

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

Subtract \frac{5}{6} from both sides of the equation.

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