Solve 6x^2=-6x+12

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

Solve for x

$$x=-2$$

$$x=1$$

Steps Using Factoring By Grouping

Add 6x to both sides.

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

Subtract 12 from both sides.

$$6x^{2}+6x-12=0$$

Divide both sides by 6.

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

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.

$$a+b=1 ab=1\left(-2\right)=-2$$

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. The only such pair is the system solution.

$$a=-1 b=2$$

Rewrite x^{2}+x-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

Add 6x to both sides.

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

Subtract 12 from both sides.

$$6x^{2}+6x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 6 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 6\left(-12\right)}}{2\times 6}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 6\left(-12\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-6±\sqrt{36-24\left(-12\right)}}{2\times 6}$$

Multiply -24 times -12.

$$x=\frac{-6±\sqrt{36+288}}{2\times 6}$$

Add 36 to 288.

$$x=\frac{-6±\sqrt{324}}{2\times 6}$$

Take the square root of 324.

$$x=\frac{-6±18}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-6±18}{12}$$

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

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

Divide 12 by 12.

$$x=1$$

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

$$x=-\frac{24}{12}$$

Divide -24 by 12.

$$x=-2$$

The equation is now solved.

$$x=1 x=-2$$

Steps for Completing the Square

Add 6x to both sides.

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

Divide both sides by 6.

$$\frac{6x^{2}+6x}{6}=\frac{12}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\frac{6}{6}x=\frac{12}{6}$$

Divide 6 by 6.

$$x^{2}+x=\frac{12}{6}$$

Divide 12 by 6.

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

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 2 to \frac{1}{4}.

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

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

$$x=1 x=-2$$