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n^{2}+3n-12-6=0
Subtract 6 from both sides.
n^{2}+3n-18=0
Subtract 6 from -12 to get -18.
a+b=3 ab=-18
To solve the equation, factor n^{2}+3n-18 using formula n^{2}+\left(a+b\right)n+ab=\left(n+a\right)\left(n+b\right). To find a and b, set up a system to be solved.
-1,18 -2,9 -3,6
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 -18.
-1+18=17 -2+9=7 -3+6=3
Calculate the sum for each pair.
a=-3 b=6
The solution is the pair that gives sum 3.
\left(n-3\right)\left(n+6\right)
Rewrite factored expression \left(n+a\right)\left(n+b\right) using the obtained values.
n=3 n=-6
To find equation solutions, solve n-3=0 and n+6=0.
n^{2}+3n-12-6=0
Subtract 6 from both sides.
n^{2}+3n-18=0
Subtract 6 from -12 to get -18.
a+b=3 ab=1\left(-18\right)=-18
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn-18. To find a and b, set up a system to be solved.
-1,18 -2,9 -3,6
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 -18.
-1+18=17 -2+9=7 -3+6=3
Calculate the sum for each pair.
a=-3 b=6
The solution is the pair that gives sum 3.
\left(n^{2}-3n\right)+\left(6n-18\right)
Rewrite n^{2}+3n-18 as \left(n^{2}-3n\right)+\left(6n-18\right).
n\left(n-3\right)+6\left(n-3\right)
Factor out n in the first and 6 in the second group.
\left(n-3\right)\left(n+6\right)
Factor out common term n-3 by using distributive property.
n=3 n=-6
To find equation solutions, solve n-3=0 and n+6=0.
n^{2}+3n-12=6
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.
n^{2}+3n-12-6=6-6
Subtract 6 from both sides of the equation.
n^{2}+3n-12-6=0
Subtracting 6 from itself leaves 0.
n^{2}+3n-18=0
Subtract 6 from -12.
n=\frac{-3±\sqrt{3^{2}-4\left(-18\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
n=\frac{-3±\sqrt{9-4\left(-18\right)}}{2}
Square 3.
n=\frac{-3±\sqrt{9+72}}{2}
Multiply -4 times -18.
n=\frac{-3±\sqrt{81}}{2}
Add 9 to 72.
n=\frac{-3±9}{2}
Take the square root of 81.
n=\frac{6}{2}
Now solve the equation n=\frac{-3±9}{2} when ± is plus. Add -3 to 9.
n=3
Divide 6 by 2.
n=-\frac{12}{2}
Now solve the equation n=\frac{-3±9}{2} when ± is minus. Subtract 9 from -3.
n=-6
Divide -12 by 2.
n=3 n=-6
The equation is now solved.
n^{2}+3n-12=6
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.
n^{2}+3n-12-\left(-12\right)=6-\left(-12\right)
Add 12 to both sides of the equation.
n^{2}+3n=6-\left(-12\right)
Subtracting -12 from itself leaves 0.
n^{2}+3n=18
Subtract -12 from 6.
n^{2}+3n+\left(\frac{3}{2}\right)^{2}=18+\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.
n^{2}+3n+\frac{9}{4}=18+\frac{9}{4}
Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.
n^{2}+3n+\frac{9}{4}=\frac{81}{4}
Add 18 to \frac{9}{4}.
\left(n+\frac{3}{2}\right)^{2}=\frac{81}{4}
Factor n^{2}+3n+\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(n+\frac{3}{2}\right)^{2}}=\sqrt{\frac{81}{4}}
Take the square root of both sides of the equation.
n+\frac{3}{2}=\frac{9}{2} n+\frac{3}{2}=-\frac{9}{2}
Simplify.
n=3 n=-6
Subtract \frac{3}{2} from both sides of the equation.