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a+b=1 ab=-650
To solve the equation, factor x^{2}+x-650 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.
-1,650 -2,325 -5,130 -10,65 -13,50 -25,26
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 -650.
-1+650=649 -2+325=323 -5+130=125 -10+65=55 -13+50=37 -25+26=1
Calculate the sum for each pair.
a=-25 b=26
The solution is the pair that gives sum 1.
\left(x-25\right)\left(x+26\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=25 x=-26
To find equation solutions, solve x-25=0 and x+26=0.
a+b=1 ab=1\left(-650\right)=-650
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-650. To find a and b, set up a system to be solved.
-1,650 -2,325 -5,130 -10,65 -13,50 -25,26
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 -650.
-1+650=649 -2+325=323 -5+130=125 -10+65=55 -13+50=37 -25+26=1
Calculate the sum for each pair.
a=-25 b=26
The solution is the pair that gives sum 1.
\left(x^{2}-25x\right)+\left(26x-650\right)
Rewrite x^{2}+x-650 as \left(x^{2}-25x\right)+\left(26x-650\right).
x\left(x-25\right)+26\left(x-25\right)
Factor out x in the first and 26 in the second group.
\left(x-25\right)\left(x+26\right)
Factor out common term x-25 by using distributive property.
x=25 x=-26
To find equation solutions, solve x-25=0 and x+26=0.
x^{2}+x-650=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{-1±\sqrt{1^{2}-4\left(-650\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -650 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-1±\sqrt{1-4\left(-650\right)}}{2}
Square 1.
x=\frac{-1±\sqrt{1+2600}}{2}
Multiply -4 times -650.
x=\frac{-1±\sqrt{2601}}{2}
Add 1 to 2600.
x=\frac{-1±51}{2}
Take the square root of 2601.
x=\frac{50}{2}
Now solve the equation x=\frac{-1±51}{2} when ± is plus. Add -1 to 51.
x=25
Divide 50 by 2.
x=-\frac{52}{2}
Now solve the equation x=\frac{-1±51}{2} when ± is minus. Subtract 51 from -1.
x=-26
Divide -52 by 2.
x=25 x=-26
The equation is now solved.
x^{2}+x-650=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}+x-650-\left(-650\right)=-\left(-650\right)
Add 650 to both sides of the equation.
x^{2}+x=-\left(-650\right)
Subtracting -650 from itself leaves 0.
x^{2}+x=650
Subtract -650 from 0.
x^{2}+x+\left(\frac{1}{2}\right)^{2}=650+\left(\frac{1}{2}\right)^{2}
Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+x+\frac{1}{4}=650+\frac{1}{4}
Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+x+\frac{1}{4}=\frac{2601}{4}
Add 650 to \frac{1}{4}.
\left(x+\frac{1}{2}\right)^{2}=\frac{2601}{4}
Factor x^{2}+x+\frac{1}{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{1}{2}\right)^{2}}=\sqrt{\frac{2601}{4}}
Take the square root of both sides of the equation.
x+\frac{1}{2}=\frac{51}{2} x+\frac{1}{2}=-\frac{51}{2}
Simplify.
x=25 x=-26
Subtract \frac{1}{2} from both sides of the equation.