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a+b=-1 ab=-6
To solve the equation, factor x^{2}-x-6 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,-6 2,-3
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 -6.
1-6=-5 2-3=-1
Calculate the sum for each pair.
a=-3 b=2
The solution is the pair that gives sum -1.
\left(x-3\right)\left(x+2\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=3 x=-2
To find equation solutions, solve x-3=0 and x+2=0.
a+b=-1 ab=1\left(-6\right)=-6
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-6. To find a and b, set up a system to be solved.
1,-6 2,-3
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 -6.
1-6=-5 2-3=-1
Calculate the sum for each pair.
a=-3 b=2
The solution is the pair that gives sum -1.
\left(x^{2}-3x\right)+\left(2x-6\right)
Rewrite x^{2}-x-6 as \left(x^{2}-3x\right)+\left(2x-6\right).
x\left(x-3\right)+2\left(x-3\right)
Factor out x in the first and 2 in the second group.
\left(x-3\right)\left(x+2\right)
Factor out common term x-3 by using distributive property.
x=3 x=-2
To find equation solutions, solve x-3=0 and x+2=0.
x^{2}-x-6=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(-1\right)±\sqrt{1-4\left(-6\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-1\right)±\sqrt{1+24}}{2}
Multiply -4 times -6.
x=\frac{-\left(-1\right)±\sqrt{25}}{2}
Add 1 to 24.
x=\frac{-\left(-1\right)±5}{2}
Take the square root of 25.
x=\frac{1±5}{2}
The opposite of -1 is 1.
x=\frac{6}{2}
Now solve the equation x=\frac{1±5}{2} when ± is plus. Add 1 to 5.
x=3
Divide 6 by 2.
x=-\frac{4}{2}
Now solve the equation x=\frac{1±5}{2} when ± is minus. Subtract 5 from 1.
x=-2
Divide -4 by 2.
x=3 x=-2
The equation is now solved.
x^{2}-x-6=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-6-\left(-6\right)=-\left(-6\right)
Add 6 to both sides of the equation.
x^{2}-x=-\left(-6\right)
Subtracting -6 from itself leaves 0.
x^{2}-x=6
Subtract -6 from 0.
x^{2}-x+\left(-\frac{1}{2}\right)^{2}=6+\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}=6+\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{25}{4}
Add 6 to \frac{1}{4}.
\left(x-\frac{1}{2}\right)^{2}=\frac{25}{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{25}{4}}
Take the square root of both sides of the equation.
x-\frac{1}{2}=\frac{5}{2} x-\frac{1}{2}=-\frac{5}{2}
Simplify.
x=3 x=-2
Add \frac{1}{2} to both sides of the equation.