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n^{2}-n-90=0
Subtract 90 from both sides.
a+b=-1 ab=-90
To solve the equation, factor n^{2}-n-90 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,-90 2,-45 3,-30 5,-18 6,-15 9,-10
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -90.
1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1
Calculate the sum for each pair.
a=-10 b=9
The solution is the pair that gives sum -1.
\left(n-10\right)\left(n+9\right)
Rewrite factored expression \left(n+a\right)\left(n+b\right) using the obtained values.
n=10 n=-9
To find equation solutions, solve n-10=0 and n+9=0.
n^{2}-n-90=0
Subtract 90 from both sides.
a+b=-1 ab=1\left(-90\right)=-90
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn-90. To find a and b, set up a system to be solved.
1,-90 2,-45 3,-30 5,-18 6,-15 9,-10
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -90.
1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1
Calculate the sum for each pair.
a=-10 b=9
The solution is the pair that gives sum -1.
\left(n^{2}-10n\right)+\left(9n-90\right)
Rewrite n^{2}-n-90 as \left(n^{2}-10n\right)+\left(9n-90\right).
n\left(n-10\right)+9\left(n-10\right)
Factor out n in the first and 9 in the second group.
\left(n-10\right)\left(n+9\right)
Factor out common term n-10 by using distributive property.
n=10 n=-9
To find equation solutions, solve n-10=0 and n+9=0.
n^{2}-n=90
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}-n-90=90-90
Subtract 90 from both sides of the equation.
n^{2}-n-90=0
Subtracting 90 from itself leaves 0.
n=\frac{-\left(-1\right)±\sqrt{1-4\left(-90\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and -90 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
n=\frac{-\left(-1\right)±\sqrt{1+360}}{2}
Multiply -4 times -90.
n=\frac{-\left(-1\right)±\sqrt{361}}{2}
Add 1 to 360.
n=\frac{-\left(-1\right)±19}{2}
Take the square root of 361.
n=\frac{1±19}{2}
The opposite of -1 is 1.
n=\frac{20}{2}
Now solve the equation n=\frac{1±19}{2} when ± is plus. Add 1 to 19.
n=10
Divide 20 by 2.
n=-\frac{18}{2}
Now solve the equation n=\frac{1±19}{2} when ± is minus. Subtract 19 from 1.
n=-9
Divide -18 by 2.
n=10 n=-9
The equation is now solved.
n^{2}-n=90
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+\left(-\frac{1}{2}\right)^{2}=90+\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}=90+\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{361}{4}
Add 90 to \frac{1}{4}.
\left(n-\frac{1}{2}\right)^{2}=\frac{361}{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{361}{4}}
Take the square root of both sides of the equation.
n-\frac{1}{2}=\frac{19}{2} n-\frac{1}{2}=-\frac{19}{2}
Simplify.
n=10 n=-9
Add \frac{1}{2} to both sides of the equation.