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a^{2}-9a-36=0
Subtract 36 from both sides.
a+b=-9 ab=-36
To solve the equation, factor a^{2}-9a-36 using formula a^{2}+\left(a+b\right)a+ab=\left(a+a\right)\left(a+b\right). To find a and b, set up a system to be solved.
1,-36 2,-18 3,-12 4,-9 6,-6
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 -36.
1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0
Calculate the sum for each pair.
a=-12 b=3
The solution is the pair that gives sum -9.
\left(a-12\right)\left(a+3\right)
Rewrite factored expression \left(a+a\right)\left(a+b\right) using the obtained values.
a=12 a=-3
To find equation solutions, solve a-12=0 and a+3=0.
a^{2}-9a-36=0
Subtract 36 from both sides.
a+b=-9 ab=1\left(-36\right)=-36
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as a^{2}+aa+ba-36. To find a and b, set up a system to be solved.
1,-36 2,-18 3,-12 4,-9 6,-6
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 -36.
1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0
Calculate the sum for each pair.
a=-12 b=3
The solution is the pair that gives sum -9.
\left(a^{2}-12a\right)+\left(3a-36\right)
Rewrite a^{2}-9a-36 as \left(a^{2}-12a\right)+\left(3a-36\right).
a\left(a-12\right)+3\left(a-12\right)
Factor out a in the first and 3 in the second group.
\left(a-12\right)\left(a+3\right)
Factor out common term a-12 by using distributive property.
a=12 a=-3
To find equation solutions, solve a-12=0 and a+3=0.
a^{2}-9a=36
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.
a^{2}-9a-36=36-36
Subtract 36 from both sides of the equation.
a^{2}-9a-36=0
Subtracting 36 from itself leaves 0.
a=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\left(-36\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -9 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-\left(-9\right)±\sqrt{81-4\left(-36\right)}}{2}
Square -9.
a=\frac{-\left(-9\right)±\sqrt{81+144}}{2}
Multiply -4 times -36.
a=\frac{-\left(-9\right)±\sqrt{225}}{2}
Add 81 to 144.
a=\frac{-\left(-9\right)±15}{2}
Take the square root of 225.
a=\frac{9±15}{2}
The opposite of -9 is 9.
a=\frac{24}{2}
Now solve the equation a=\frac{9±15}{2} when ± is plus. Add 9 to 15.
a=12
Divide 24 by 2.
a=-\frac{6}{2}
Now solve the equation a=\frac{9±15}{2} when ± is minus. Subtract 15 from 9.
a=-3
Divide -6 by 2.
a=12 a=-3
The equation is now solved.
a^{2}-9a=36
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.
a^{2}-9a+\left(-\frac{9}{2}\right)^{2}=36+\left(-\frac{9}{2}\right)^{2}
Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}-9a+\frac{81}{4}=36+\frac{81}{4}
Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.
a^{2}-9a+\frac{81}{4}=\frac{225}{4}
Add 36 to \frac{81}{4}.
\left(a-\frac{9}{2}\right)^{2}=\frac{225}{4}
Factor a^{2}-9a+\frac{81}{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(a-\frac{9}{2}\right)^{2}}=\sqrt{\frac{225}{4}}
Take the square root of both sides of the equation.
a-\frac{9}{2}=\frac{15}{2} a-\frac{9}{2}=-\frac{15}{2}
Simplify.
a=12 a=-3
Add \frac{9}{2} to both sides of the equation.