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5x^{2}-13x+6=0
Add 6 to both sides.
a+b=-13 ab=5\times 6=30
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx+6. To find a and b, set up a system to be solved.
-1,-30 -2,-15 -3,-10 -5,-6
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 30.
-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11
Calculate the sum for each pair.
a=-10 b=-3
The solution is the pair that gives sum -13.
\left(5x^{2}-10x\right)+\left(-3x+6\right)
Rewrite 5x^{2}-13x+6 as \left(5x^{2}-10x\right)+\left(-3x+6\right).
5x\left(x-2\right)-3\left(x-2\right)
Factor out 5x in the first and -3 in the second group.
\left(x-2\right)\left(5x-3\right)
Factor out common term x-2 by using distributive property.
x=2 x=\frac{3}{5}
To find equation solutions, solve x-2=0 and 5x-3=0.
5x^{2}-13x=-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.
5x^{2}-13x-\left(-6\right)=-6-\left(-6\right)
Add 6 to both sides of the equation.
5x^{2}-13x-\left(-6\right)=0
Subtracting -6 from itself leaves 0.
5x^{2}-13x+6=0
Subtract -6 from 0.
x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 5\times 6}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -13 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-13\right)±\sqrt{169-4\times 5\times 6}}{2\times 5}
Square -13.
x=\frac{-\left(-13\right)±\sqrt{169-20\times 6}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-13\right)±\sqrt{169-120}}{2\times 5}
Multiply -20 times 6.
x=\frac{-\left(-13\right)±\sqrt{49}}{2\times 5}
Add 169 to -120.
x=\frac{-\left(-13\right)±7}{2\times 5}
Take the square root of 49.
x=\frac{13±7}{2\times 5}
The opposite of -13 is 13.
x=\frac{13±7}{10}
Multiply 2 times 5.
x=\frac{20}{10}
Now solve the equation x=\frac{13±7}{10} when ± is plus. Add 13 to 7.
x=2
Divide 20 by 10.
x=\frac{6}{10}
Now solve the equation x=\frac{13±7}{10} when ± is minus. Subtract 7 from 13.
x=\frac{3}{5}
Reduce the fraction \frac{6}{10} to lowest terms by extracting and canceling out 2.
x=2 x=\frac{3}{5}
The equation is now solved.
5x^{2}-13x=-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.
\frac{5x^{2}-13x}{5}=-\frac{6}{5}
Divide both sides by 5.
x^{2}-\frac{13}{5}x=-\frac{6}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-\frac{13}{5}x+\left(-\frac{13}{10}\right)^{2}=-\frac{6}{5}+\left(-\frac{13}{10}\right)^{2}
Divide -\frac{13}{5}, the coefficient of the x term, by 2 to get -\frac{13}{10}. Then add the square of -\frac{13}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{13}{5}x+\frac{169}{100}=-\frac{6}{5}+\frac{169}{100}
Square -\frac{13}{10} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{13}{5}x+\frac{169}{100}=\frac{49}{100}
Add -\frac{6}{5} to \frac{169}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{13}{10}\right)^{2}=\frac{49}{100}
Factor x^{2}-\frac{13}{5}x+\frac{169}{100}. 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{13}{10}\right)^{2}}=\sqrt{\frac{49}{100}}
Take the square root of both sides of the equation.
x-\frac{13}{10}=\frac{7}{10} x-\frac{13}{10}=-\frac{7}{10}
Simplify.
x=2 x=\frac{3}{5}
Add \frac{13}{10} to both sides of the equation.