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6x^{2}-x-5=0
Subtract 5 from both sides.
a+b=-1 ab=6\left(-5\right)=-30
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx-5. To find a and b, set up a system to be solved.
1,-30 2,-15 3,-10 5,-6
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 -30.
1-30=-29 2-15=-13 3-10=-7 5-6=-1
Calculate the sum for each pair.
a=-6 b=5
The solution is the pair that gives sum -1.
\left(6x^{2}-6x\right)+\left(5x-5\right)
Rewrite 6x^{2}-x-5 as \left(6x^{2}-6x\right)+\left(5x-5\right).
6x\left(x-1\right)+5\left(x-1\right)
Factor out 6x in the first and 5 in the second group.
\left(x-1\right)\left(6x+5\right)
Factor out common term x-1 by using distributive property.
x=1 x=-\frac{5}{6}
To find equation solutions, solve x-1=0 and 6x+5=0.
6x^{2}-x=5
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.
6x^{2}-x-5=5-5
Subtract 5 from both sides of the equation.
6x^{2}-x-5=0
Subtracting 5 from itself leaves 0.
x=\frac{-\left(-1\right)±\sqrt{1-4\times 6\left(-5\right)}}{2\times 6}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -1 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-1\right)±\sqrt{1-24\left(-5\right)}}{2\times 6}
Multiply -4 times 6.
x=\frac{-\left(-1\right)±\sqrt{1+120}}{2\times 6}
Multiply -24 times -5.
x=\frac{-\left(-1\right)±\sqrt{121}}{2\times 6}
Add 1 to 120.
x=\frac{-\left(-1\right)±11}{2\times 6}
Take the square root of 121.
x=\frac{1±11}{2\times 6}
The opposite of -1 is 1.
x=\frac{1±11}{12}
Multiply 2 times 6.
x=\frac{12}{12}
Now solve the equation x=\frac{1±11}{12} when ± is plus. Add 1 to 11.
x=1
Divide 12 by 12.
x=-\frac{10}{12}
Now solve the equation x=\frac{1±11}{12} when ± is minus. Subtract 11 from 1.
x=-\frac{5}{6}
Reduce the fraction \frac{-10}{12} to lowest terms by extracting and canceling out 2.
x=1 x=-\frac{5}{6}
The equation is now solved.
6x^{2}-x=5
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.
\frac{6x^{2}-x}{6}=\frac{5}{6}
Divide both sides by 6.
x^{2}-\frac{1}{6}x=\frac{5}{6}
Dividing by 6 undoes the multiplication by 6.
x^{2}-\frac{1}{6}x+\left(-\frac{1}{12}\right)^{2}=\frac{5}{6}+\left(-\frac{1}{12}\right)^{2}
Divide -\frac{1}{6}, the coefficient of the x term, by 2 to get -\frac{1}{12}. Then add the square of -\frac{1}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{1}{6}x+\frac{1}{144}=\frac{5}{6}+\frac{1}{144}
Square -\frac{1}{12} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{1}{6}x+\frac{1}{144}=\frac{121}{144}
Add \frac{5}{6} to \frac{1}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{12}\right)^{2}=\frac{121}{144}
Factor x^{2}-\frac{1}{6}x+\frac{1}{144}. 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}{12}\right)^{2}}=\sqrt{\frac{121}{144}}
Take the square root of both sides of the equation.
x-\frac{1}{12}=\frac{11}{12} x-\frac{1}{12}=-\frac{11}{12}
Simplify.
x=1 x=-\frac{5}{6}
Add \frac{1}{12} to both sides of the equation.