Solve 2x^2+36=3x^2+x-20

$$2 x ^ { 2 } + 36 = 3 x ^ { 2 } + x - 20$$

Solve for x

$$x=-8$$

$$x=7$$

Steps Using Factoring By Grouping

Subtract 3x^{2} from both sides.

$$2x^{2}+36-3x^{2}=x-20$$

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

$$-x^{2}+36=x-20$$

Subtract x from both sides.

$$-x^{2}+36-x=-20$$

Add 20 to both sides.

$$-x^{2}+36-x+20=0$$

Add 36 and 20 to get 56.

$$-x^{2}+56-x=0$$

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

$$-x^{2}-x+56=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+56. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-56=-56$$

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 -56.

$$1,-56 2,-28 4,-14 7,-8$$

Calculate the sum for each pair.

$$1-56=-55 2-28=-26 4-14=-10 7-8=-1$$

The solution is the pair that gives sum -1.

$$a=7 b=-8$$

Rewrite -x^{2}-x+56 as \left(-x^{2}+7x\right)+\left(-8x+56\right).

$$\left(-x^{2}+7x\right)+\left(-8x+56\right)$$

Factor out x in the first and 8 in the second group.

$$x\left(-x+7\right)+8\left(-x+7\right)$$

Factor out common term -x+7 by using distributive property.

$$\left(-x+7\right)\left(x+8\right)$$

To find equation solutions, solve -x+7=0 and x+8=0.

$$x=7 x=-8$$

Steps Using the Quadratic Formula

Subtract 3x^{2} from both sides.

$$2x^{2}+36-3x^{2}=x-20$$

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

$$-x^{2}+36=x-20$$

Subtract x from both sides.

$$-x^{2}+36-x=-20$$

Add 20 to both sides.

$$-x^{2}+36-x+20=0$$

Add 36 and 20 to get 56.

$$-x^{2}+56-x=0$$

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.

$$-x^{2}-x+56=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -1 for b, and 56 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-1\right)\times 56}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-1\right)±\sqrt{1+4\times 56}}{2\left(-1\right)}$$

Multiply 4 times 56.

$$x=\frac{-\left(-1\right)±\sqrt{1+224}}{2\left(-1\right)}$$

Add 1 to 224.

$$x=\frac{-\left(-1\right)±\sqrt{225}}{2\left(-1\right)}$$

Take the square root of 225.

$$x=\frac{-\left(-1\right)±15}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±15}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±15}{-2}$$

Now solve the equation x=\frac{1±15}{-2} when ± is plus. Add 1 to 15.

$$x=\frac{16}{-2}$$

Divide 16 by -2.

$$x=-8$$

Now solve the equation x=\frac{1±15}{-2} when ± is minus. Subtract 15 from 1.

$$x=-\frac{14}{-2}$$

Divide -14 by -2.

$$x=7$$

The equation is now solved.

$$x=-8 x=7$$

Steps for Completing the Square

Subtract 3x^{2} from both sides.

$$2x^{2}+36-3x^{2}=x-20$$

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

$$-x^{2}+36=x-20$$

Subtract x from both sides.

$$-x^{2}+36-x=-20$$

Subtract 36 from both sides.

$$-x^{2}-x=-20-36$$

Subtract 36 from -20 to get -56.

$$-x^{2}-x=-56$$

Divide both sides by -1.

$$\frac{-x^{2}-x}{-1}=-\frac{56}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{1}{-1}\right)x=-\frac{56}{-1}$$

Divide -1 by -1.

$$x^{2}+x=-\frac{56}{-1}$$

Divide -56 by -1.

$$x^{2}+x=56$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=56+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=56+\frac{1}{4}$$

Add 56 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{225}{4}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{15}{2} x+\frac{1}{2}=-\frac{15}{2}$$

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

$$x=7 x=-8$$