Solve for m
m=-13
m=10
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m^{2}+3m-130=0
Subtract 130 from both sides.
a+b=3 ab=-130
To solve the equation, factor m^{2}+3m-130 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.
-1,130 -2,65 -5,26 -10,13
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 -130.
-1+130=129 -2+65=63 -5+26=21 -10+13=3
Calculate the sum for each pair.
a=-10 b=13
The solution is the pair that gives sum 3.
\left(m-10\right)\left(m+13\right)
Rewrite factored expression \left(m+a\right)\left(m+b\right) using the obtained values.
m=10 m=-13
To find equation solutions, solve m-10=0 and m+13=0.
m^{2}+3m-130=0
Subtract 130 from both sides.
a+b=3 ab=1\left(-130\right)=-130
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as m^{2}+am+bm-130. To find a and b, set up a system to be solved.
-1,130 -2,65 -5,26 -10,13
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 -130.
-1+130=129 -2+65=63 -5+26=21 -10+13=3
Calculate the sum for each pair.
a=-10 b=13
The solution is the pair that gives sum 3.
\left(m^{2}-10m\right)+\left(13m-130\right)
Rewrite m^{2}+3m-130 as \left(m^{2}-10m\right)+\left(13m-130\right).
m\left(m-10\right)+13\left(m-10\right)
Factor out m in the first and 13 in the second group.
\left(m-10\right)\left(m+13\right)
Factor out common term m-10 by using distributive property.
m=10 m=-13
To find equation solutions, solve m-10=0 and m+13=0.
m^{2}+3m=130
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.
m^{2}+3m-130=130-130
Subtract 130 from both sides of the equation.
m^{2}+3m-130=0
Subtracting 130 from itself leaves 0.
m=\frac{-3±\sqrt{3^{2}-4\left(-130\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -130 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
m=\frac{-3±\sqrt{9-4\left(-130\right)}}{2}
Square 3.
m=\frac{-3±\sqrt{9+520}}{2}
Multiply -4 times -130.
m=\frac{-3±\sqrt{529}}{2}
Add 9 to 520.
m=\frac{-3±23}{2}
Take the square root of 529.
m=\frac{20}{2}
Now solve the equation m=\frac{-3±23}{2} when ± is plus. Add -3 to 23.
m=10
Divide 20 by 2.
m=-\frac{26}{2}
Now solve the equation m=\frac{-3±23}{2} when ± is minus. Subtract 23 from -3.
m=-13
Divide -26 by 2.
m=10 m=-13
The equation is now solved.
m^{2}+3m=130
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.
m^{2}+3m+\left(\frac{3}{2}\right)^{2}=130+\left(\frac{3}{2}\right)^{2}
Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
m^{2}+3m+\frac{9}{4}=130+\frac{9}{4}
Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.
m^{2}+3m+\frac{9}{4}=\frac{529}{4}
Add 130 to \frac{9}{4}.
\left(m+\frac{3}{2}\right)^{2}=\frac{529}{4}
Factor m^{2}+3m+\frac{9}{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}.
\sqrt{\left(m+\frac{3}{2}\right)^{2}}=\sqrt{\frac{529}{4}}
Take the square root of both sides of the equation.
m+\frac{3}{2}=\frac{23}{2} m+\frac{3}{2}=-\frac{23}{2}
Simplify.
m=10 m=-13
Subtract \frac{3}{2} from both sides of the equation.
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