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3x^{2}-13x-10=0
Subtract 10 from both sides.
a+b=-13 ab=3\left(-10\right)=-30
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-10. 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=-15 b=2
The solution is the pair that gives sum -13.
\left(3x^{2}-15x\right)+\left(2x-10\right)
Rewrite 3x^{2}-13x-10 as \left(3x^{2}-15x\right)+\left(2x-10\right).
3x\left(x-5\right)+2\left(x-5\right)
Factor out 3x in the first and 2 in the second group.
\left(x-5\right)\left(3x+2\right)
Factor out common term x-5 by using distributive property.
x=5 x=-\frac{2}{3}
To find equation solutions, solve x-5=0 and 3x+2=0.
3x^{2}-13x=10
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.
3x^{2}-13x-10=10-10
Subtract 10 from both sides of the equation.
3x^{2}-13x-10=0
Subtracting 10 from itself leaves 0.
x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 3\left(-10\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -13 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-13\right)±\sqrt{169-4\times 3\left(-10\right)}}{2\times 3}
Square -13.
x=\frac{-\left(-13\right)±\sqrt{169-12\left(-10\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-13\right)±\sqrt{169+120}}{2\times 3}
Multiply -12 times -10.
x=\frac{-\left(-13\right)±\sqrt{289}}{2\times 3}
Add 169 to 120.
x=\frac{-\left(-13\right)±17}{2\times 3}
Take the square root of 289.
x=\frac{13±17}{2\times 3}
The opposite of -13 is 13.
x=\frac{13±17}{6}
Multiply 2 times 3.
x=\frac{30}{6}
Now solve the equation x=\frac{13±17}{6} when ± is plus. Add 13 to 17.
x=5
Divide 30 by 6.
x=-\frac{4}{6}
Now solve the equation x=\frac{13±17}{6} when ± is minus. Subtract 17 from 13.
x=-\frac{2}{3}
Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.
x=5 x=-\frac{2}{3}
The equation is now solved.
3x^{2}-13x=10
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{3x^{2}-13x}{3}=\frac{10}{3}
Divide both sides by 3.
x^{2}-\frac{13}{3}x=\frac{10}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-\frac{13}{3}x+\left(-\frac{13}{6}\right)^{2}=\frac{10}{3}+\left(-\frac{13}{6}\right)^{2}
Divide -\frac{13}{3}, the coefficient of the x term, by 2 to get -\frac{13}{6}. Then add the square of -\frac{13}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{13}{3}x+\frac{169}{36}=\frac{10}{3}+\frac{169}{36}
Square -\frac{13}{6} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{13}{3}x+\frac{169}{36}=\frac{289}{36}
Add \frac{10}{3} to \frac{169}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{13}{6}\right)^{2}=\frac{289}{36}
Factor x^{2}-\frac{13}{3}x+\frac{169}{36}. 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}{6}\right)^{2}}=\sqrt{\frac{289}{36}}
Take the square root of both sides of the equation.
x-\frac{13}{6}=\frac{17}{6} x-\frac{13}{6}=-\frac{17}{6}
Simplify.
x=5 x=-\frac{2}{3}
Add \frac{13}{6} to both sides of the equation.