Solve -11x=2x^2+12

$$- 11 x = 2 x ^ { 2 } + 12$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=-4$$

Steps Using Factoring By Grouping

Subtract 2x^{2} from both sides.

$$-11x-2x^{2}=12$$

Subtract 12 from both sides.

$$-11x-2x^{2}-12=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-2x^{2}-11x-12=0$$

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

$$a+b=-11 ab=-2\left(-12\right)=24$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -11.

$$a=-3 b=-8$$

Rewrite -2x^{2}-11x-12 as \left(-2x^{2}-3x\right)+\left(-8x-12\right).

$$\left(-2x^{2}-3x\right)+\left(-8x-12\right)$$

Factor out -x in the first and -4 in the second group.

$$-x\left(2x+3\right)-4\left(2x+3\right)$$

Factor out common term 2x+3 by using distributive property.

$$\left(2x+3\right)\left(-x-4\right)$$

To find equation solutions, solve 2x+3=0 and -x-4=0.

$$x=-\frac{3}{2} x=-4$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

$$-11x-2x^{2}=12$$

Subtract 12 from both sides.

$$-11x-2x^{2}-12=0$$

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.

$$-2x^{2}-11x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -11 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-2\right)\left(-12\right)}}{2\left(-2\right)}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\left(-2\right)\left(-12\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-11\right)±\sqrt{121+8\left(-12\right)}}{2\left(-2\right)}$$

Multiply 8 times -12.

$$x=\frac{-\left(-11\right)±\sqrt{121-96}}{2\left(-2\right)}$$

Add 121 to -96.

$$x=\frac{-\left(-11\right)±\sqrt{25}}{2\left(-2\right)}$$

Take the square root of 25.

$$x=\frac{-\left(-11\right)±5}{2\left(-2\right)}$$

The opposite of -11 is 11.

$$x=\frac{11±5}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{11±5}{-4}$$

Now solve the equation x=\frac{11±5}{-4} when ± is plus. Add 11 to 5.

$$x=\frac{16}{-4}$$

Divide 16 by -4.

$$x=-4$$

Now solve the equation x=\frac{11±5}{-4} when ± is minus. Subtract 5 from 11.

$$x=\frac{6}{-4}$$

Reduce the fraction \frac{6}{-4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=-4 x=-\frac{3}{2}$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

$$-11x-2x^{2}=12$$

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.

$$-2x^{2}-11x=12$$

Divide both sides by -2.

$$\frac{-2x^{2}-11x}{-2}=\frac{12}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{11}{-2}\right)x=\frac{12}{-2}$$

Divide -11 by -2.

$$x^{2}+\frac{11}{2}x=\frac{12}{-2}$$

Divide 12 by -2.

$$x^{2}+\frac{11}{2}x=-6$$

Divide \frac{11}{2}, the coefficient of the x term, by 2 to get \frac{11}{4}. Then add the square of \frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{2}x+\left(\frac{11}{4}\right)^{2}=-6+\left(\frac{11}{4}\right)^{2}$$

Square \frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{2}x+\frac{121}{16}=-6+\frac{121}{16}$$

Add -6 to \frac{121}{16}.

$$x^{2}+\frac{11}{2}x+\frac{121}{16}=\frac{25}{16}$$

Factor x^{2}+\frac{11}{2}x+\frac{121}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{11}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x+\frac{11}{4}=\frac{5}{4} x+\frac{11}{4}=-\frac{5}{4}$$

Subtract \frac{11}{4} from both sides of the equation.

$$x=-\frac{3}{2} x=-4$$