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x^{2}+5x-104=0
Subtract 104 from both sides.
a+b=5 ab=-104
To solve the equation, factor x^{2}+5x-104 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.
-1,104 -2,52 -4,26 -8,13
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -104.
-1+104=103 -2+52=50 -4+26=22 -8+13=5
Calculate the sum for each pair.
a=-8 b=13
The solution is the pair that gives sum 5.
\left(x-8\right)\left(x+13\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=8 x=-13
To find equation solutions, solve x-8=0 and x+13=0.
x^{2}+5x-104=0
Subtract 104 from both sides.
a+b=5 ab=1\left(-104\right)=-104
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-104. To find a and b, set up a system to be solved.
-1,104 -2,52 -4,26 -8,13
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -104.
-1+104=103 -2+52=50 -4+26=22 -8+13=5
Calculate the sum for each pair.
a=-8 b=13
The solution is the pair that gives sum 5.
\left(x^{2}-8x\right)+\left(13x-104\right)
Rewrite x^{2}+5x-104 as \left(x^{2}-8x\right)+\left(13x-104\right).
x\left(x-8\right)+13\left(x-8\right)
Factor out x in the first and 13 in the second group.
\left(x-8\right)\left(x+13\right)
Factor out common term x-8 by using distributive property.
x=8 x=-13
To find equation solutions, solve x-8=0 and x+13=0.
x^{2}+5x=104
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^{2}+5x-104=104-104
Subtract 104 from both sides of the equation.
x^{2}+5x-104=0
Subtracting 104 from itself leaves 0.
x=\frac{-5±\sqrt{5^{2}-4\left(-104\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and -104 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-5±\sqrt{25-4\left(-104\right)}}{2}
Square 5.
x=\frac{-5±\sqrt{25+416}}{2}
Multiply -4 times -104.
x=\frac{-5±\sqrt{441}}{2}
Add 25 to 416.
x=\frac{-5±21}{2}
Take the square root of 441.
x=\frac{16}{2}
Now solve the equation x=\frac{-5±21}{2} when ± is plus. Add -5 to 21.
x=8
Divide 16 by 2.
x=-\frac{26}{2}
Now solve the equation x=\frac{-5±21}{2} when ± is minus. Subtract 21 from -5.
x=-13
Divide -26 by 2.
x=8 x=-13
The equation is now solved.
x^{2}+5x=104
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}+5x+\left(\frac{5}{2}\right)^{2}=104+\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.
x^{2}+5x+\frac{25}{4}=104+\frac{25}{4}
Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+5x+\frac{25}{4}=\frac{441}{4}
Add 104 to \frac{25}{4}.
\left(x+\frac{5}{2}\right)^{2}=\frac{441}{4}
Factor x^{2}+5x+\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(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{441}{4}}
Take the square root of both sides of the equation.
x+\frac{5}{2}=\frac{21}{2} x+\frac{5}{2}=-\frac{21}{2}
Simplify.
x=8 x=-13
Subtract \frac{5}{2} from both sides of the equation.