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a+b=-124 ab=-512
To solve the equation, factor x^{2}-124x-512 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.
1,-512 2,-256 4,-128 8,-64 16,-32
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 -512.
1-512=-511 2-256=-254 4-128=-124 8-64=-56 16-32=-16
Calculate the sum for each pair.
a=-128 b=4
The solution is the pair that gives sum -124.
\left(x-128\right)\left(x+4\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=128 x=-4
To find equation solutions, solve x-128=0 and x+4=0.
a+b=-124 ab=1\left(-512\right)=-512
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-512. To find a and b, set up a system to be solved.
1,-512 2,-256 4,-128 8,-64 16,-32
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 -512.
1-512=-511 2-256=-254 4-128=-124 8-64=-56 16-32=-16
Calculate the sum for each pair.
a=-128 b=4
The solution is the pair that gives sum -124.
\left(x^{2}-128x\right)+\left(4x-512\right)
Rewrite x^{2}-124x-512 as \left(x^{2}-128x\right)+\left(4x-512\right).
x\left(x-128\right)+4\left(x-128\right)
Factor out x in the first and 4 in the second group.
\left(x-128\right)\left(x+4\right)
Factor out common term x-128 by using distributive property.
x=128 x=-4
To find equation solutions, solve x-128=0 and x+4=0.
x^{2}-124x-512=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.
x=\frac{-\left(-124\right)±\sqrt{\left(-124\right)^{2}-4\left(-512\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -124 for b, and -512 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-124\right)±\sqrt{15376-4\left(-512\right)}}{2}
Square -124.
x=\frac{-\left(-124\right)±\sqrt{15376+2048}}{2}
Multiply -4 times -512.
x=\frac{-\left(-124\right)±\sqrt{17424}}{2}
Add 15376 to 2048.
x=\frac{-\left(-124\right)±132}{2}
Take the square root of 17424.
x=\frac{124±132}{2}
The opposite of -124 is 124.
x=\frac{256}{2}
Now solve the equation x=\frac{124±132}{2} when ± is plus. Add 124 to 132.
x=128
Divide 256 by 2.
x=-\frac{8}{2}
Now solve the equation x=\frac{124±132}{2} when ± is minus. Subtract 132 from 124.
x=-4
Divide -8 by 2.
x=128 x=-4
The equation is now solved.
x^{2}-124x-512=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.
x^{2}-124x-512-\left(-512\right)=-\left(-512\right)
Add 512 to both sides of the equation.
x^{2}-124x=-\left(-512\right)
Subtracting -512 from itself leaves 0.
x^{2}-124x=512
Subtract -512 from 0.
x^{2}-124x+\left(-62\right)^{2}=512+\left(-62\right)^{2}
Divide -124, the coefficient of the x term, by 2 to get -62. Then add the square of -62 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-124x+3844=512+3844
Square -62.
x^{2}-124x+3844=4356
Add 512 to 3844.
\left(x-62\right)^{2}=4356
Factor x^{2}-124x+3844. 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(x-62\right)^{2}}=\sqrt{4356}
Take the square root of both sides of the equation.
x-62=66 x-62=-66
Simplify.
x=128 x=-4
Add 62 to both sides of the equation.