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x^{2}-7x-60=0
Subtract 60 from both sides.
a+b=-7 ab=-60
To solve the equation, factor x^{2}-7x-60 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,-60 2,-30 3,-20 4,-15 5,-12 6,-10
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 -60.
1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4
Calculate the sum for each pair.
a=-12 b=5
The solution is the pair that gives sum -7.
\left(x-12\right)\left(x+5\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=12 x=-5
To find equation solutions, solve x-12=0 and x+5=0.
x^{2}-7x-60=0
Subtract 60 from both sides.
a+b=-7 ab=1\left(-60\right)=-60
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-60. To find a and b, set up a system to be solved.
1,-60 2,-30 3,-20 4,-15 5,-12 6,-10
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 -60.
1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4
Calculate the sum for each pair.
a=-12 b=5
The solution is the pair that gives sum -7.
\left(x^{2}-12x\right)+\left(5x-60\right)
Rewrite x^{2}-7x-60 as \left(x^{2}-12x\right)+\left(5x-60\right).
x\left(x-12\right)+5\left(x-12\right)
Factor out x in the first and 5 in the second group.
\left(x-12\right)\left(x+5\right)
Factor out common term x-12 by using distributive property.
x=12 x=-5
To find equation solutions, solve x-12=0 and x+5=0.
x^{2}-7x=60
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}-7x-60=60-60
Subtract 60 from both sides of the equation.
x^{2}-7x-60=0
Subtracting 60 from itself leaves 0.
x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-60\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and -60 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-7\right)±\sqrt{49-4\left(-60\right)}}{2}
Square -7.
x=\frac{-\left(-7\right)±\sqrt{49+240}}{2}
Multiply -4 times -60.
x=\frac{-\left(-7\right)±\sqrt{289}}{2}
Add 49 to 240.
x=\frac{-\left(-7\right)±17}{2}
Take the square root of 289.
x=\frac{7±17}{2}
The opposite of -7 is 7.
x=\frac{24}{2}
Now solve the equation x=\frac{7±17}{2} when ± is plus. Add 7 to 17.
x=12
Divide 24 by 2.
x=-\frac{10}{2}
Now solve the equation x=\frac{7±17}{2} when ± is minus. Subtract 17 from 7.
x=-5
Divide -10 by 2.
x=12 x=-5
The equation is now solved.
x^{2}-7x=60
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}-7x+\left(-\frac{7}{2}\right)^{2}=60+\left(-\frac{7}{2}\right)^{2}
Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-7x+\frac{49}{4}=60+\frac{49}{4}
Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-7x+\frac{49}{4}=\frac{289}{4}
Add 60 to \frac{49}{4}.
\left(x-\frac{7}{2}\right)^{2}=\frac{289}{4}
Factor x^{2}-7x+\frac{49}{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{7}{2}\right)^{2}}=\sqrt{\frac{289}{4}}
Take the square root of both sides of the equation.
x-\frac{7}{2}=\frac{17}{2} x-\frac{7}{2}=-\frac{17}{2}
Simplify.
x=12 x=-5
Add \frac{7}{2} to both sides of the equation.