Solve m^2=-2m+120

$$m ^ { 2 } = - 2 m + 120$$

Solve for m

$$m=-12$$

$$m=10$$

Steps Using Factoring

Add 2m to both sides.

$$m^{2}+2m=120$$

Subtract 120 from both sides.

$$m^{2}+2m-120=0$$

To solve the equation, factor m^{2}+2m-120 using formula m^{2}+\left(a+b\right)m+ab=\left(m+a\right)\left(m+b\right). To find a and b, set up a system to be solved.

$$a+b=2 ab=-120$$

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. List all such integer pairs that give product -120.

$$-1,120 -2,60 -3,40 -4,30 -5,24 -6,20 -8,15 -10,12$$

Calculate the sum for each pair.

$$-1+120=119 -2+60=58 -3+40=37 -4+30=26 -5+24=19 -6+20=14 -8+15=7 -10+12=2$$

The solution is the pair that gives sum 2.

$$a=-10 b=12$$

Rewrite factored expression \left(m+a\right)\left(m+b\right) using the obtained values.

$$\left(m-10\right)\left(m+12\right)$$

To find equation solutions, solve m-10=0 and m+12=0.

$$m=10 m=-12$$

Steps Using Factoring By Grouping

Add 2m to both sides.

$$m^{2}+2m=120$$

Subtract 120 from both sides.

$$m^{2}+2m-120=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as m^{2}+am+bm-120. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-120\right)=-120$$

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. List all such integer pairs that give product -120.

$$-1,120 -2,60 -3,40 -4,30 -5,24 -6,20 -8,15 -10,12$$

Calculate the sum for each pair.

$$-1+120=119 -2+60=58 -3+40=37 -4+30=26 -5+24=19 -6+20=14 -8+15=7 -10+12=2$$

The solution is the pair that gives sum 2.

$$a=-10 b=12$$

Rewrite m^{2}+2m-120 as \left(m^{2}-10m\right)+\left(12m-120\right).

$$\left(m^{2}-10m\right)+\left(12m-120\right)$$

Factor out m in the first and 12 in the second group.

$$m\left(m-10\right)+12\left(m-10\right)$$

Factor out common term m-10 by using distributive property.

$$\left(m-10\right)\left(m+12\right)$$

To find equation solutions, solve m-10=0 and m+12=0.

$$m=10 m=-12$$

Steps Using the Quadratic Formula

Add 2m to both sides.

$$m^{2}+2m=120$$

Subtract 120 from both sides.

$$m^{2}+2m-120=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -120 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-2±\sqrt{2^{2}-4\left(-120\right)}}{2}$$

Square 2.

$$m=\frac{-2±\sqrt{4-4\left(-120\right)}}{2}$$

Multiply -4 times -120.

$$m=\frac{-2±\sqrt{4+480}}{2}$$

Add 4 to 480.

$$m=\frac{-2±\sqrt{484}}{2}$$

Take the square root of 484.

$$m=\frac{-2±22}{2}$$

Now solve the equation m=\frac{-2±22}{2} when ± is plus. Add -2 to 22.

$$m=\frac{20}{2}$$

Divide 20 by 2.

$$m=10$$

Now solve the equation m=\frac{-2±22}{2} when ± is minus. Subtract 22 from -2.

$$m=-\frac{24}{2}$$

Divide -24 by 2.

$$m=-12$$

The equation is now solved.

$$m=10 m=-12$$

Steps for Completing the Square

Add 2m to both sides.

$$m^{2}+2m=120$$

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.

$$m^{2}+2m+1^{2}=120+1^{2}$$

Square 1.

$$m^{2}+2m+1=120+1$$

Add 120 to 1.

$$m^{2}+2m+1=121$$

Factor m^{2}+2m+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}.

$$\left(m+1\right)^{2}=121$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m+1\right)^{2}}=\sqrt{121}$$

Simplify.

$$m+1=11 m+1=-11$$

Subtract 1 from both sides of the equation.

$$m=10 m=-12$$