Solve 3x^2-2x-8=5x^2-20

$$3 x ^ { 2 } - 2 x - 8 = 5 x ^ { 2 } - 20$$

Solve for x

$$x=-3$$

$$x=2$$

Steps Using Factoring By Grouping

Subtract 5x^{2} from both sides.

$$3x^{2}-2x-8-5x^{2}=-20$$

Combine 3x^{2} and -5x^{2} to get -2x^{2}.

$$-2x^{2}-2x-8=-20$$

Add 20 to both sides.

$$-2x^{2}-2x-8+20=0$$

Add -8 and 20 to get 12.

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

Divide both sides by 2.

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

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

$$a+b=-1 ab=-6=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=2 b=-3$$

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

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

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

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

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

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

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

$$x=2 x=-3$$

Steps Using the Quadratic Formula

Subtract 5x^{2} from both sides.

$$3x^{2}-2x-8-5x^{2}=-20$$

Combine 3x^{2} and -5x^{2} to get -2x^{2}.

$$-2x^{2}-2x-8=-20$$

Add 20 to both sides.

$$-2x^{2}-2x-8+20=0$$

Add -8 and 20 to get 12.

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

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

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

Square -2.

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

Multiply -4 times -2.

$$x=\frac{-\left(-2\right)±\sqrt{4+8\times 12}}{2\left(-2\right)}$$

Multiply 8 times 12.

$$x=\frac{-\left(-2\right)±\sqrt{4+96}}{2\left(-2\right)}$$

Add 4 to 96.

$$x=\frac{-\left(-2\right)±\sqrt{100}}{2\left(-2\right)}$$

Take the square root of 100.

$$x=\frac{-\left(-2\right)±10}{2\left(-2\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±10}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{2±10}{-4}$$

Now solve the equation x=\frac{2±10}{-4} when ± is plus. Add 2 to 10.

$$x=\frac{12}{-4}$$

Divide 12 by -4.

$$x=-3$$

Now solve the equation x=\frac{2±10}{-4} when ± is minus. Subtract 10 from 2.

$$x=-\frac{8}{-4}$$

Divide -8 by -4.

$$x=2$$

The equation is now solved.

$$x=-3 x=2$$

Steps for Completing the Square

Subtract 5x^{2} from both sides.

$$3x^{2}-2x-8-5x^{2}=-20$$

Combine 3x^{2} and -5x^{2} to get -2x^{2}.

$$-2x^{2}-2x-8=-20$$

Add 8 to both sides.

$$-2x^{2}-2x=-20+8$$

Add -20 and 8 to get -12.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -2 by -2.

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

Divide -12 by -2.

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

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.

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

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

$$x^{2}+x+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{25}{4}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{5}{2} x+\frac{1}{2}=-\frac{5}{2}$$

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

$$x=2 x=-3$$