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5x^{2}-8-18x=0
Subtract 18x from both sides.
5x^{2}-18x-8=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=-18 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=-20 b=2
The solution is the pair that gives sum -18.
\left(5x^{2}-20x\right)+\left(2x-8\right)
Rewrite 5x^{2}-18x-8 as \left(5x^{2}-20x\right)+\left(2x-8\right).
5x\left(x-4\right)+2\left(x-4\right)
Factor out 5x in the first and 2 in the second group.
\left(x-4\right)\left(5x+2\right)
Factor out common term x-4 by using distributive property.
x=4 x=-\frac{2}{5}
To find equation solutions, solve x-4=0 and 5x+2=0.
5x^{2}-8-18x=0
Subtract 18x from both sides.
5x^{2}-18x-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(-18\right)±\sqrt{\left(-18\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, -18 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-18\right)±\sqrt{324-4\times 5\left(-8\right)}}{2\times 5}
Square -18.
x=\frac{-\left(-18\right)±\sqrt{324-20\left(-8\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-18\right)±\sqrt{324+160}}{2\times 5}
Multiply -20 times -8.
x=\frac{-\left(-18\right)±\sqrt{484}}{2\times 5}
Add 324 to 160.
x=\frac{-\left(-18\right)±22}{2\times 5}
Take the square root of 484.
x=\frac{18±22}{2\times 5}
The opposite of -18 is 18.
x=\frac{18±22}{10}
Multiply 2 times 5.
x=\frac{40}{10}
Now solve the equation x=\frac{18±22}{10} when ± is plus. Add 18 to 22.
x=4
Divide 40 by 10.
x=-\frac{4}{10}
Now solve the equation x=\frac{18±22}{10} when ± is minus. Subtract 22 from 18.
x=-\frac{2}{5}
Reduce the fraction \frac{-4}{10} to lowest terms by extracting and canceling out 2.
x=4 x=-\frac{2}{5}
The equation is now solved.
5x^{2}-8-18x=0
Subtract 18x from both sides.
5x^{2}-18x=8
Add 8 to both sides. Anything plus zero gives itself.
\frac{5x^{2}-18x}{5}=\frac{8}{5}
Divide both sides by 5.
x^{2}-\frac{18}{5}x=\frac{8}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-\frac{18}{5}x+\left(-\frac{9}{5}\right)^{2}=\frac{8}{5}+\left(-\frac{9}{5}\right)^{2}
Divide -\frac{18}{5}, the coefficient of the x term, by 2 to get -\frac{9}{5}. Then add the square of -\frac{9}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{18}{5}x+\frac{81}{25}=\frac{8}{5}+\frac{81}{25}
Square -\frac{9}{5} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{18}{5}x+\frac{81}{25}=\frac{121}{25}
Add \frac{8}{5} to \frac{81}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{9}{5}\right)^{2}=\frac{121}{25}
Factor x^{2}-\frac{18}{5}x+\frac{81}{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{9}{5}\right)^{2}}=\sqrt{\frac{121}{25}}
Take the square root of both sides of the equation.
x-\frac{9}{5}=\frac{11}{5} x-\frac{9}{5}=-\frac{11}{5}
Simplify.
x=4 x=-\frac{2}{5}
Add \frac{9}{5} to both sides of the equation.