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x^{2}-5x-14=0
Subtract 14 from both sides.
a+b=-5 ab=-14
To solve the equation, factor x^{2}-5x-14 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,-14 2,-7
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 -14.
1-14=-13 2-7=-5
Calculate the sum for each pair.
a=-7 b=2
The solution is the pair that gives sum -5.
\left(x-7\right)\left(x+2\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=7 x=-2
To find equation solutions, solve x-7=0 and x+2=0.
x^{2}-5x-14=0
Subtract 14 from both sides.
a+b=-5 ab=1\left(-14\right)=-14
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-14. To find a and b, set up a system to be solved.
1,-14 2,-7
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 -14.
1-14=-13 2-7=-5
Calculate the sum for each pair.
a=-7 b=2
The solution is the pair that gives sum -5.
\left(x^{2}-7x\right)+\left(2x-14\right)
Rewrite x^{2}-5x-14 as \left(x^{2}-7x\right)+\left(2x-14\right).
x\left(x-7\right)+2\left(x-7\right)
Factor out x in the first and 2 in the second group.
\left(x-7\right)\left(x+2\right)
Factor out common term x-7 by using distributive property.
x=7 x=-2
To find equation solutions, solve x-7=0 and x+2=0.
x^{2}-5x=14
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-14=14-14
Subtract 14 from both sides of the equation.
x^{2}-5x-14=0
Subtracting 14 from itself leaves 0.
x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-14\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and -14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-5\right)±\sqrt{25-4\left(-14\right)}}{2}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25+56}}{2}
Multiply -4 times -14.
x=\frac{-\left(-5\right)±\sqrt{81}}{2}
Add 25 to 56.
x=\frac{-\left(-5\right)±9}{2}
Take the square root of 81.
x=\frac{5±9}{2}
The opposite of -5 is 5.
x=\frac{14}{2}
Now solve the equation x=\frac{5±9}{2} when ± is plus. Add 5 to 9.
x=7
Divide 14 by 2.
x=-\frac{4}{2}
Now solve the equation x=\frac{5±9}{2} when ± is minus. Subtract 9 from 5.
x=-2
Divide -4 by 2.
x=7 x=-2
The equation is now solved.
x^{2}-5x=14
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}=14+\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}=14+\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{81}{4}
Add 14 to \frac{25}{4}.
\left(x-\frac{5}{2}\right)^{2}=\frac{81}{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{81}{4}}
Take the square root of both sides of the equation.
x-\frac{5}{2}=\frac{9}{2} x-\frac{5}{2}=-\frac{9}{2}
Simplify.
x=7 x=-2
Add \frac{5}{2} to both sides of the equation.