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3x^{2}-10x-8=0
Subtract 8 from both sides.
a+b=-10 ab=3\left(-8\right)=-24
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-8. To find a and b, set up a system to be solved.
1,-24 2,-12 3,-8 4,-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 -24.
1-24=-23 2-12=-10 3-8=-5 4-6=-2
Calculate the sum for each pair.
a=-12 b=2
The solution is the pair that gives sum -10.
\left(3x^{2}-12x\right)+\left(2x-8\right)
Rewrite 3x^{2}-10x-8 as \left(3x^{2}-12x\right)+\left(2x-8\right).
3x\left(x-4\right)+2\left(x-4\right)
Factor out 3x in the first and 2 in the second group.
\left(x-4\right)\left(3x+2\right)
Factor out common term x-4 by using distributive property.
x=4 x=-\frac{2}{3}
To find equation solutions, solve x-4=0 and 3x+2=0.
3x^{2}-10x=8
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}-10x-8=8-8
Subtract 8 from both sides of the equation.
3x^{2}-10x-8=0
Subtracting 8 from itself leaves 0.
x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 3\left(-8\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -10 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-10\right)±\sqrt{100-4\times 3\left(-8\right)}}{2\times 3}
Square -10.
x=\frac{-\left(-10\right)±\sqrt{100-12\left(-8\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-10\right)±\sqrt{100+96}}{2\times 3}
Multiply -12 times -8.
x=\frac{-\left(-10\right)±\sqrt{196}}{2\times 3}
Add 100 to 96.
x=\frac{-\left(-10\right)±14}{2\times 3}
Take the square root of 196.
x=\frac{10±14}{2\times 3}
The opposite of -10 is 10.
x=\frac{10±14}{6}
Multiply 2 times 3.
x=\frac{24}{6}
Now solve the equation x=\frac{10±14}{6} when ± is plus. Add 10 to 14.
x=4
Divide 24 by 6.
x=-\frac{4}{6}
Now solve the equation x=\frac{10±14}{6} when ± is minus. Subtract 14 from 10.
x=-\frac{2}{3}
Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.
x=4 x=-\frac{2}{3}
The equation is now solved.
3x^{2}-10x=8
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}-10x}{3}=\frac{8}{3}
Divide both sides by 3.
x^{2}-\frac{10}{3}x=\frac{8}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=\frac{8}{3}+\left(-\frac{5}{3}\right)^{2}
Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{8}{3}+\frac{25}{9}
Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{49}{9}
Add \frac{8}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{5}{3}\right)^{2}=\frac{49}{9}
Factor x^{2}-\frac{10}{3}x+\frac{25}{9}. 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}{3}\right)^{2}}=\sqrt{\frac{49}{9}}
Take the square root of both sides of the equation.
x-\frac{5}{3}=\frac{7}{3} x-\frac{5}{3}=-\frac{7}{3}
Simplify.
x=4 x=-\frac{2}{3}
Add \frac{5}{3} to both sides of the equation.