Solve 12x^2+5x-27=3x^2-13x

$$12 x ^ { 2 } + 5 x - 27 = 3 x ^ { 2 } - 13 x$$

Solve for x

$$x=-3$$

$$x=1$$

Steps Using Factoring By Grouping

Subtract 3x^{2} from both sides.

$$12x^{2}+5x-27-3x^{2}=-13x$$

Combine 12x^{2} and -3x^{2} to get 9x^{2}.

$$9x^{2}+5x-27=-13x$$

Add 13x to both sides.

$$9x^{2}+5x-27+13x=0$$

Combine 5x and 13x to get 18x.

$$9x^{2}+18x-27=0$$

Divide both sides by 9.

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

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

$$a+b=2 ab=1\left(-3\right)=-3$$

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.

$$a=-1 b=3$$

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

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

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

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

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

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

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

$$x=1 x=-3$$

Steps Using the Quadratic Formula

Subtract 3x^{2} from both sides.

$$12x^{2}+5x-27-3x^{2}=-13x$$

Combine 12x^{2} and -3x^{2} to get 9x^{2}.

$$9x^{2}+5x-27=-13x$$

Add 13x to both sides.

$$9x^{2}+5x-27+13x=0$$

Combine 5x and 13x to get 18x.

$$9x^{2}+18x-27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 18 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\times 9\left(-27\right)}}{2\times 9}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\times 9\left(-27\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-18±\sqrt{324-36\left(-27\right)}}{2\times 9}$$

Multiply -36 times -27.

$$x=\frac{-18±\sqrt{324+972}}{2\times 9}$$

Add 324 to 972.

$$x=\frac{-18±\sqrt{1296}}{2\times 9}$$

Take the square root of 1296.

$$x=\frac{-18±36}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-18±36}{18}$$

Now solve the equation x=\frac{-18±36}{18} when ± is plus. Add -18 to 36.

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

Divide 18 by 18.

$$x=1$$

Now solve the equation x=\frac{-18±36}{18} when ± is minus. Subtract 36 from -18.

$$x=-\frac{54}{18}$$

Divide -54 by 18.

$$x=-3$$

The equation is now solved.

$$x=1 x=-3$$

Steps for Completing the Square

Subtract 3x^{2} from both sides.

$$12x^{2}+5x-27-3x^{2}=-13x$$

Combine 12x^{2} and -3x^{2} to get 9x^{2}.

$$9x^{2}+5x-27=-13x$$

Add 13x to both sides.

$$9x^{2}+5x-27+13x=0$$

Combine 5x and 13x to get 18x.

$$9x^{2}+18x-27=0$$

Add 27 to both sides. Anything plus zero gives itself.

$$9x^{2}+18x=27$$

Divide both sides by 9.

$$\frac{9x^{2}+18x}{9}=\frac{27}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{18}{9}x=\frac{27}{9}$$

Divide 18 by 9.

$$x^{2}+2x=\frac{27}{9}$$

Divide 27 by 9.

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

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

$$x^{2}+2x+1^{2}=3+1^{2}$$

Square 1.

$$x^{2}+2x+1=3+1$$

Add 3 to 1.

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

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 1 from both sides of the equation.

$$x=1 x=-3$$