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x^{2}+5x-6=0
Subtract 6 from both sides.
a+b=5 ab=-6
To solve the equation, factor x^{2}+5x-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 positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.
-1+6=5 -2+3=1
Calculate the sum for each pair.
a=-1 b=6
The solution is the pair that gives sum 5.
\left(x-1\right)\left(x+6\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=1 x=-6
To find equation solutions, solve x-1=0 and x+6=0.
x^{2}+5x-6=0
Subtract 6 from both sides.
a+b=5 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 positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.
-1+6=5 -2+3=1
Calculate the sum for each pair.
a=-1 b=6
The solution is the pair that gives sum 5.
\left(x^{2}-x\right)+\left(6x-6\right)
Rewrite x^{2}+5x-6 as \left(x^{2}-x\right)+\left(6x-6\right).
x\left(x-1\right)+6\left(x-1\right)
Factor out x in the first and 6 in the second group.
\left(x-1\right)\left(x+6\right)
Factor out common term x-1 by using distributive property.
x=1 x=-6
To find equation solutions, solve x-1=0 and x+6=0.
x^{2}+5x=6
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-6=6-6
Subtract 6 from both sides of the equation.
x^{2}+5x-6=0
Subtracting 6 from itself leaves 0.
x=\frac{-5±\sqrt{5^{2}-4\left(-6\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-5±\sqrt{25-4\left(-6\right)}}{2}
Square 5.
x=\frac{-5±\sqrt{25+24}}{2}
Multiply -4 times -6.
x=\frac{-5±\sqrt{49}}{2}
Add 25 to 24.
x=\frac{-5±7}{2}
Take the square root of 49.
x=\frac{2}{2}
Now solve the equation x=\frac{-5±7}{2} when ± is plus. Add -5 to 7.
x=1
Divide 2 by 2.
x=-\frac{12}{2}
Now solve the equation x=\frac{-5±7}{2} when ± is minus. Subtract 7 from -5.
x=-6
Divide -12 by 2.
x=1 x=-6
The equation is now solved.
x^{2}+5x=6
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}=6+\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}=6+\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{49}{4}
Add 6 to \frac{25}{4}.
\left(x+\frac{5}{2}\right)^{2}=\frac{49}{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{49}{4}}
Take the square root of both sides of the equation.
x+\frac{5}{2}=\frac{7}{2} x+\frac{5}{2}=-\frac{7}{2}
Simplify.
x=1 x=-6
Subtract \frac{5}{2} from both sides of the equation.