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r^{2}-r-36=4r
Subtract 36 from both sides.
r^{2}-r-36-4r=0
Subtract 4r from both sides.
r^{2}-5r-36=0
Combine -r and -4r to get -5r.
a+b=-5 ab=-36
To solve the equation, factor r^{2}-5r-36 using formula r^{2}+\left(a+b\right)r+ab=\left(r+a\right)\left(r+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=-9 b=4
The solution is the pair that gives sum -5.
\left(r-9\right)\left(r+4\right)
Rewrite factored expression \left(r+a\right)\left(r+b\right) using the obtained values.
r=9 r=-4
To find equation solutions, solve r-9=0 and r+4=0.
r^{2}-r-36=4r
Subtract 36 from both sides.
r^{2}-r-36-4r=0
Subtract 4r from both sides.
r^{2}-5r-36=0
Combine -r and -4r to get -5r.
a+b=-5 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 r^{2}+ar+br-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=-9 b=4
The solution is the pair that gives sum -5.
\left(r^{2}-9r\right)+\left(4r-36\right)
Rewrite r^{2}-5r-36 as \left(r^{2}-9r\right)+\left(4r-36\right).
r\left(r-9\right)+4\left(r-9\right)
Factor out r in the first and 4 in the second group.
\left(r-9\right)\left(r+4\right)
Factor out common term r-9 by using distributive property.
r=9 r=-4
To find equation solutions, solve r-9=0 and r+4=0.
r^{2}-r-36=4r
Subtract 36 from both sides.
r^{2}-r-36-4r=0
Subtract 4r from both sides.
r^{2}-5r-36=0
Combine -r and -4r to get -5r.
r=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-36\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
r=\frac{-\left(-5\right)±\sqrt{25-4\left(-36\right)}}{2}
Square -5.
r=\frac{-\left(-5\right)±\sqrt{25+144}}{2}
Multiply -4 times -36.
r=\frac{-\left(-5\right)±\sqrt{169}}{2}
Add 25 to 144.
r=\frac{-\left(-5\right)±13}{2}
Take the square root of 169.
r=\frac{5±13}{2}
The opposite of -5 is 5.
r=\frac{18}{2}
Now solve the equation r=\frac{5±13}{2} when ± is plus. Add 5 to 13.
r=9
Divide 18 by 2.
r=-\frac{8}{2}
Now solve the equation r=\frac{5±13}{2} when ± is minus. Subtract 13 from 5.
r=-4
Divide -8 by 2.
r=9 r=-4
The equation is now solved.
r^{2}-r-4r=36
Subtract 4r from both sides.
r^{2}-5r=36
Combine -r and -4r to get -5r.
r^{2}-5r+\left(-\frac{5}{2}\right)^{2}=36+\left(-\frac{5}{2}\right)^{2}
Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
r^{2}-5r+\frac{25}{4}=36+\frac{25}{4}
Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.
r^{2}-5r+\frac{25}{4}=\frac{169}{4}
Add 36 to \frac{25}{4}.
\left(r-\frac{5}{2}\right)^{2}=\frac{169}{4}
Factor r^{2}-5r+\frac{25}{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(r-\frac{5}{2}\right)^{2}}=\sqrt{\frac{169}{4}}
Take the square root of both sides of the equation.
r-\frac{5}{2}=\frac{13}{2} r-\frac{5}{2}=-\frac{13}{2}
Simplify.
r=9 r=-4
Add \frac{5}{2} to both sides of the equation.