Solve for m
m=-4
m=2
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3m^{2}+6m-31+7=0
Add 7 to both sides.
3m^{2}+6m-24=0
Add -31 and 7 to get -24.
m^{2}+2m-8=0
Divide both sides by 3.
a+b=2 ab=1\left(-8\right)=-8
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as m^{2}+am+bm-8. To find a and b, set up a system to be solved.
-1,8 -2,4
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 -8.
-1+8=7 -2+4=2
Calculate the sum for each pair.
a=-2 b=4
The solution is the pair that gives sum 2.
\left(m^{2}-2m\right)+\left(4m-8\right)
Rewrite m^{2}+2m-8 as \left(m^{2}-2m\right)+\left(4m-8\right).
m\left(m-2\right)+4\left(m-2\right)
Factor out m in the first and 4 in the second group.
\left(m-2\right)\left(m+4\right)
Factor out common term m-2 by using distributive property.
m=2 m=-4
To find equation solutions, solve m-2=0 and m+4=0.
3m^{2}+6m-31=-7
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.
3m^{2}+6m-31-\left(-7\right)=-7-\left(-7\right)
Add 7 to both sides of the equation.
3m^{2}+6m-31-\left(-7\right)=0
Subtracting -7 from itself leaves 0.
3m^{2}+6m-24=0
Subtract -7 from -31.
m=\frac{-6±\sqrt{6^{2}-4\times 3\left(-24\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 6 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
m=\frac{-6±\sqrt{36-4\times 3\left(-24\right)}}{2\times 3}
Square 6.
m=\frac{-6±\sqrt{36-12\left(-24\right)}}{2\times 3}
Multiply -4 times 3.
m=\frac{-6±\sqrt{36+288}}{2\times 3}
Multiply -12 times -24.
m=\frac{-6±\sqrt{324}}{2\times 3}
Add 36 to 288.
m=\frac{-6±18}{2\times 3}
Take the square root of 324.
m=\frac{-6±18}{6}
Multiply 2 times 3.
m=\frac{12}{6}
Now solve the equation m=\frac{-6±18}{6} when ± is plus. Add -6 to 18.
m=2
Divide 12 by 6.
m=-\frac{24}{6}
Now solve the equation m=\frac{-6±18}{6} when ± is minus. Subtract 18 from -6.
m=-4
Divide -24 by 6.
m=2 m=-4
The equation is now solved.
3m^{2}+6m-31=-7
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.
3m^{2}+6m-31-\left(-31\right)=-7-\left(-31\right)
Add 31 to both sides of the equation.
3m^{2}+6m=-7-\left(-31\right)
Subtracting -31 from itself leaves 0.
3m^{2}+6m=24
Subtract -31 from -7.
\frac{3m^{2}+6m}{3}=\frac{24}{3}
Divide both sides by 3.
m^{2}+\frac{6}{3}m=\frac{24}{3}
Dividing by 3 undoes the multiplication by 3.
m^{2}+2m=\frac{24}{3}
Divide 6 by 3.
m^{2}+2m=8
Divide 24 by 3.
m^{2}+2m+1^{2}=8+1^{2}
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=8+1
Square 1.
m^{2}+2m+1=9
Add 8 to 1.
\left(m+1\right)^{2}=9
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}.
\sqrt{\left(m+1\right)^{2}}=\sqrt{9}
Take the square root of both sides of the equation.
m+1=3 m+1=-3
Simplify.
m=2 m=-4
Subtract 1 from both sides of the equation.
Examples
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{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}