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a+b=-6 ab=5\left(-8\right)=-40
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-8. To find a and b, set up a system to be solved.
1,-40 2,-20 4,-10 5,-8
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 -40.
1-40=-39 2-20=-18 4-10=-6 5-8=-3
Calculate the sum for each pair.
a=-10 b=4
The solution is the pair that gives sum -6.
\left(5x^{2}-10x\right)+\left(4x-8\right)
Rewrite 5x^{2}-6x-8 as \left(5x^{2}-10x\right)+\left(4x-8\right).
5x\left(x-2\right)+4\left(x-2\right)
Factor out 5x in the first and 4 in the second group.
\left(x-2\right)\left(5x+4\right)
Factor out common term x-2 by using distributive property.
x=2 x=-\frac{4}{5}
To find equation solutions, solve x-2=0 and 5x+4=0.
5x^{2}-6x-8=0
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=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 5\left(-8\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -6 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-6\right)±\sqrt{36-4\times 5\left(-8\right)}}{2\times 5}
Square -6.
x=\frac{-\left(-6\right)±\sqrt{36-20\left(-8\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-6\right)±\sqrt{36+160}}{2\times 5}
Multiply -20 times -8.
x=\frac{-\left(-6\right)±\sqrt{196}}{2\times 5}
Add 36 to 160.
x=\frac{-\left(-6\right)±14}{2\times 5}
Take the square root of 196.
x=\frac{6±14}{2\times 5}
The opposite of -6 is 6.
x=\frac{6±14}{10}
Multiply 2 times 5.
x=\frac{20}{10}
Now solve the equation x=\frac{6±14}{10} when ± is plus. Add 6 to 14.
x=2
Divide 20 by 10.
x=-\frac{8}{10}
Now solve the equation x=\frac{6±14}{10} when ± is minus. Subtract 14 from 6.
x=-\frac{4}{5}
Reduce the fraction \frac{-8}{10} to lowest terms by extracting and canceling out 2.
x=2 x=-\frac{4}{5}
The equation is now solved.
5x^{2}-6x-8=0
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.
5x^{2}-6x-8-\left(-8\right)=-\left(-8\right)
Add 8 to both sides of the equation.
5x^{2}-6x=-\left(-8\right)
Subtracting -8 from itself leaves 0.
5x^{2}-6x=8
Subtract -8 from 0.
\frac{5x^{2}-6x}{5}=\frac{8}{5}
Divide both sides by 5.
x^{2}-\frac{6}{5}x=\frac{8}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-\frac{6}{5}x+\left(-\frac{3}{5}\right)^{2}=\frac{8}{5}+\left(-\frac{3}{5}\right)^{2}
Divide -\frac{6}{5}, the coefficient of the x term, by 2 to get -\frac{3}{5}. Then add the square of -\frac{3}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{6}{5}x+\frac{9}{25}=\frac{8}{5}+\frac{9}{25}
Square -\frac{3}{5} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{6}{5}x+\frac{9}{25}=\frac{49}{25}
Add \frac{8}{5} to \frac{9}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{3}{5}\right)^{2}=\frac{49}{25}
Factor x^{2}-\frac{6}{5}x+\frac{9}{25}. 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{3}{5}\right)^{2}}=\sqrt{\frac{49}{25}}
Take the square root of both sides of the equation.
x-\frac{3}{5}=\frac{7}{5} x-\frac{3}{5}=-\frac{7}{5}
Simplify.
x=2 x=-\frac{4}{5}
Add \frac{3}{5} to both sides of the equation.