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-16b+b^{2}-225=0
Subtract 225 from both sides.
b^{2}-16b-225=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=-16 ab=-225
To solve the equation, factor b^{2}-16b-225 using formula b^{2}+\left(a+b\right)b+ab=\left(b+a\right)\left(b+b\right). To find a and b, set up a system to be solved.
1,-225 3,-75 5,-45 9,-25 15,-15
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 -225.
1-225=-224 3-75=-72 5-45=-40 9-25=-16 15-15=0
Calculate the sum for each pair.
a=-25 b=9
The solution is the pair that gives sum -16.
\left(b-25\right)\left(b+9\right)
Rewrite factored expression \left(b+a\right)\left(b+b\right) using the obtained values.
b=25 b=-9
To find equation solutions, solve b-25=0 and b+9=0.
-16b+b^{2}-225=0
Subtract 225 from both sides.
b^{2}-16b-225=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=-16 ab=1\left(-225\right)=-225
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as b^{2}+ab+bb-225. To find a and b, set up a system to be solved.
1,-225 3,-75 5,-45 9,-25 15,-15
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 -225.
1-225=-224 3-75=-72 5-45=-40 9-25=-16 15-15=0
Calculate the sum for each pair.
a=-25 b=9
The solution is the pair that gives sum -16.
\left(b^{2}-25b\right)+\left(9b-225\right)
Rewrite b^{2}-16b-225 as \left(b^{2}-25b\right)+\left(9b-225\right).
b\left(b-25\right)+9\left(b-25\right)
Factor out b in the first and 9 in the second group.
\left(b-25\right)\left(b+9\right)
Factor out common term b-25 by using distributive property.
b=25 b=-9
To find equation solutions, solve b-25=0 and b+9=0.
b^{2}-16b=225
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.
b^{2}-16b-225=225-225
Subtract 225 from both sides of the equation.
b^{2}-16b-225=0
Subtracting 225 from itself leaves 0.
b=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\left(-225\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -16 for b, and -225 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
b=\frac{-\left(-16\right)±\sqrt{256-4\left(-225\right)}}{2}
Square -16.
b=\frac{-\left(-16\right)±\sqrt{256+900}}{2}
Multiply -4 times -225.
b=\frac{-\left(-16\right)±\sqrt{1156}}{2}
Add 256 to 900.
b=\frac{-\left(-16\right)±34}{2}
Take the square root of 1156.
b=\frac{16±34}{2}
The opposite of -16 is 16.
b=\frac{50}{2}
Now solve the equation b=\frac{16±34}{2} when ± is plus. Add 16 to 34.
b=25
Divide 50 by 2.
b=-\frac{18}{2}
Now solve the equation b=\frac{16±34}{2} when ± is minus. Subtract 34 from 16.
b=-9
Divide -18 by 2.
b=25 b=-9
The equation is now solved.
b^{2}-16b=225
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.
b^{2}-16b+\left(-8\right)^{2}=225+\left(-8\right)^{2}
Divide -16, the coefficient of the x term, by 2 to get -8. Then add the square of -8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
b^{2}-16b+64=225+64
Square -8.
b^{2}-16b+64=289
Add 225 to 64.
\left(b-8\right)^{2}=289
Factor b^{2}-16b+64. 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(b-8\right)^{2}}=\sqrt{289}
Take the square root of both sides of the equation.
b-8=17 b-8=-17
Simplify.
b=25 b=-9
Add 8 to both sides of the equation.