Solve for n
n=-9
n=6
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n^{2}+3n=54
Multiply both sides of the equation by 2.
n^{2}+3n-54=0
Subtract 54 from both sides.
a+b=3 ab=-54
To solve the equation, factor n^{2}+3n-54 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.
-1,54 -2,27 -3,18 -6,9
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 -54.
-1+54=53 -2+27=25 -3+18=15 -6+9=3
Calculate the sum for each pair.
a=-6 b=9
The solution is the pair that gives sum 3.
\left(n-6\right)\left(n+9\right)
Rewrite factored expression \left(n+a\right)\left(n+b\right) using the obtained values.
n=6 n=-9
To find equation solutions, solve n-6=0 and n+9=0.
n^{2}+3n=54
Multiply both sides of the equation by 2.
n^{2}+3n-54=0
Subtract 54 from both sides.
a+b=3 ab=1\left(-54\right)=-54
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn-54. To find a and b, set up a system to be solved.
-1,54 -2,27 -3,18 -6,9
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 -54.
-1+54=53 -2+27=25 -3+18=15 -6+9=3
Calculate the sum for each pair.
a=-6 b=9
The solution is the pair that gives sum 3.
\left(n^{2}-6n\right)+\left(9n-54\right)
Rewrite n^{2}+3n-54 as \left(n^{2}-6n\right)+\left(9n-54\right).
n\left(n-6\right)+9\left(n-6\right)
Factor out n in the first and 9 in the second group.
\left(n-6\right)\left(n+9\right)
Factor out common term n-6 by using distributive property.
n=6 n=-9
To find equation solutions, solve n-6=0 and n+9=0.
n^{2}+3n=54
Multiply both sides of the equation by 2.
n^{2}+3n-54=0
Subtract 54 from both sides.
n=\frac{-3±\sqrt{3^{2}-4\left(-54\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -54 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
n=\frac{-3±\sqrt{9-4\left(-54\right)}}{2}
Square 3.
n=\frac{-3±\sqrt{9+216}}{2}
Multiply -4 times -54.
n=\frac{-3±\sqrt{225}}{2}
Add 9 to 216.
n=\frac{-3±15}{2}
Take the square root of 225.
n=\frac{12}{2}
Now solve the equation n=\frac{-3±15}{2} when ± is plus. Add -3 to 15.
n=6
Divide 12 by 2.
n=-\frac{18}{2}
Now solve the equation n=\frac{-3±15}{2} when ± is minus. Subtract 15 from -3.
n=-9
Divide -18 by 2.
n=6 n=-9
The equation is now solved.
n^{2}+3n=54
Multiply both sides of the equation by 2.
n^{2}+3n+\left(\frac{3}{2}\right)^{2}=54+\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}=54+\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{225}{4}
Add 54 to \frac{9}{4}.
\left(n+\frac{3}{2}\right)^{2}=\frac{225}{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{225}{4}}
Take the square root of both sides of the equation.
n+\frac{3}{2}=\frac{15}{2} n+\frac{3}{2}=-\frac{15}{2}
Simplify.
n=6 n=-9
Subtract \frac{3}{2} from both sides of the equation.
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Limits
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