Solve 6x^2+12x=5x-2

$$6 x ^ { 2 } + 12 x = 5 x - 2$$

Solve for x

$$x=-\frac{2}{3}\approx -0.666666667$$

$$x=-\frac{1}{2}=-0.5$$

Steps Using Factoring By Grouping

Subtract 5x from both sides.

$$6x^{2}+12x-5x=-2$$

Combine 12x and -5x to get 7x.

$$6x^{2}+7x=-2$$

Add 2 to both sides.

$$6x^{2}+7x+2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=7 ab=6\times 2=12$$

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 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=3 b=4$$

Rewrite 6x^{2}+7x+2 as \left(6x^{2}+3x\right)+\left(4x+2\right).

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

Factor out 3x in the first and 2 in the second group.

$$3x\left(2x+1\right)+2\left(2x+1\right)$$

Factor out common term 2x+1 by using distributive property.

$$\left(2x+1\right)\left(3x+2\right)$$

To find equation solutions, solve 2x+1=0 and 3x+2=0.

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

Steps Using the Quadratic Formula

Subtract 5x from both sides.

$$6x^{2}+12x-5x=-2$$

Combine 12x and -5x to get 7x.

$$6x^{2}+7x=-2$$

Add 2 to both sides.

$$6x^{2}+7x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 7 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 6\times 2}}{2\times 6}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 6\times 2}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-7±\sqrt{49-24\times 2}}{2\times 6}$$

Multiply -24 times 2.

$$x=\frac{-7±\sqrt{49-48}}{2\times 6}$$

Add 49 to -48.

$$x=\frac{-7±\sqrt{1}}{2\times 6}$$

Take the square root of 1.

$$x=\frac{-7±1}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-7±1}{12}$$

Now solve the equation x=\frac{-7±1}{12} when ± is plus. Add -7 to 1.

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

Reduce the fraction \frac{-6}{12} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{-7±1}{12} when ± is minus. Subtract 1 from -7.

$$x=-\frac{8}{12}$$

Reduce the fraction \frac{-8}{12} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 5x from both sides.

$$6x^{2}+12x-5x=-2$$

Combine 12x and -5x to get 7x.

$$6x^{2}+7x=-2$$

Divide both sides by 6.

$$\frac{6x^{2}+7x}{6}=-\frac{2}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\frac{7}{6}x=-\frac{2}{6}$$

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

$$x^{2}+\frac{7}{6}x=-\frac{1}{3}$$

Divide \frac{7}{6}, the coefficient of the x term, by 2 to get \frac{7}{12}. Then add the square of \frac{7}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{6}x+\left(\frac{7}{12}\right)^{2}=-\frac{1}{3}+\left(\frac{7}{12}\right)^{2}$$

Square \frac{7}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{6}x+\frac{49}{144}=-\frac{1}{3}+\frac{49}{144}$$

Add -\frac{1}{3} to \frac{49}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{6}x+\frac{49}{144}=\frac{1}{144}$$

Factor x^{2}+\frac{7}{6}x+\frac{49}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{12}\right)^{2}=\frac{1}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{12}\right)^{2}}=\sqrt{\frac{1}{144}}$$

Simplify.

$$x+\frac{7}{12}=\frac{1}{12} x+\frac{7}{12}=-\frac{1}{12}$$

Subtract \frac{7}{12} from both sides of the equation.

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