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a+b=1 ab=-72
To solve the equation, factor n^{2}+n-72 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.
-1,72 -2,36 -3,24 -4,18 -6,12 -8,9
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 -72.
-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1
Calculate the sum for each pair.
a=-8 b=9
The solution is the pair that gives sum 1.
\left(n-8\right)\left(n+9\right)
Rewrite factored expression \left(n+a\right)\left(n+b\right) using the obtained values.
n=8 n=-9
To find equation solutions, solve n-8=0 and n+9=0.
a+b=1 ab=1\left(-72\right)=-72
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn-72. To find a and b, set up a system to be solved.
-1,72 -2,36 -3,24 -4,18 -6,12 -8,9
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 -72.
-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1
Calculate the sum for each pair.
a=-8 b=9
The solution is the pair that gives sum 1.
\left(n^{2}-8n\right)+\left(9n-72\right)
Rewrite n^{2}+n-72 as \left(n^{2}-8n\right)+\left(9n-72\right).
n\left(n-8\right)+9\left(n-8\right)
Factor out n in the first and 9 in the second group.
\left(n-8\right)\left(n+9\right)
Factor out common term n-8 by using distributive property.
n=8 n=-9
To find equation solutions, solve n-8=0 and n+9=0.
n^{2}+n-72=0
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=\frac{-1±\sqrt{1^{2}-4\left(-72\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
n=\frac{-1±\sqrt{1-4\left(-72\right)}}{2}
Square 1.
n=\frac{-1±\sqrt{1+288}}{2}
Multiply -4 times -72.
n=\frac{-1±\sqrt{289}}{2}
Add 1 to 288.
n=\frac{-1±17}{2}
Take the square root of 289.
n=\frac{16}{2}
Now solve the equation n=\frac{-1±17}{2} when ± is plus. Add -1 to 17.
n=8
Divide 16 by 2.
n=-\frac{18}{2}
Now solve the equation n=\frac{-1±17}{2} when ± is minus. Subtract 17 from -1.
n=-9
Divide -18 by 2.
n=8 n=-9
The equation is now solved.
n^{2}+n-72=0
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}+n-72-\left(-72\right)=-\left(-72\right)
Add 72 to both sides of the equation.
n^{2}+n=-\left(-72\right)
Subtracting -72 from itself leaves 0.
n^{2}+n=72
Subtract -72 from 0.
n^{2}+n+\left(\frac{1}{2}\right)^{2}=72+\left(\frac{1}{2}\right)^{2}
Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
n^{2}+n+\frac{1}{4}=72+\frac{1}{4}
Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.
n^{2}+n+\frac{1}{4}=\frac{289}{4}
Add 72 to \frac{1}{4}.
\left(n+\frac{1}{2}\right)^{2}=\frac{289}{4}
Factor n^{2}+n+\frac{1}{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{1}{2}\right)^{2}}=\sqrt{\frac{289}{4}}
Take the square root of both sides of the equation.
n+\frac{1}{2}=\frac{17}{2} n+\frac{1}{2}=-\frac{17}{2}
Simplify.
n=8 n=-9
Subtract \frac{1}{2} from both sides of the equation.