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-\frac{40}{27}x^{2}+\frac{160}{3}x-30=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{-\frac{160}{3}±\sqrt{\left(\frac{160}{3}\right)^{2}-4\left(-\frac{40}{27}\right)\left(-30\right)}}{2\left(-\frac{40}{27}\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{40}{27} for a, \frac{160}{3} for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\frac{160}{3}±\sqrt{\frac{25600}{9}-4\left(-\frac{40}{27}\right)\left(-30\right)}}{2\left(-\frac{40}{27}\right)}
Square \frac{160}{3} by squaring both the numerator and the denominator of the fraction.
x=\frac{-\frac{160}{3}±\sqrt{\frac{25600}{9}+\frac{160}{27}\left(-30\right)}}{2\left(-\frac{40}{27}\right)}
Multiply -4 times -\frac{40}{27}.
x=\frac{-\frac{160}{3}±\sqrt{\frac{25600-1600}{9}}}{2\left(-\frac{40}{27}\right)}
Multiply \frac{160}{27} times -30.
x=\frac{-\frac{160}{3}±\sqrt{\frac{8000}{3}}}{2\left(-\frac{40}{27}\right)}
Add \frac{25600}{9} to -\frac{1600}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=\frac{-\frac{160}{3}±\frac{40\sqrt{15}}{3}}{2\left(-\frac{40}{27}\right)}
Take the square root of \frac{8000}{3}.
x=\frac{-\frac{160}{3}±\frac{40\sqrt{15}}{3}}{-\frac{80}{27}}
Multiply 2 times -\frac{40}{27}.
x=\frac{40\sqrt{15}-160}{-\frac{80}{27}\times 3}
Now solve the equation x=\frac{-\frac{160}{3}±\frac{40\sqrt{15}}{3}}{-\frac{80}{27}} when ± is plus. Add -\frac{160}{3} to \frac{40\sqrt{15}}{3}.
x=-\frac{9\sqrt{15}}{2}+18
Divide \frac{-160+40\sqrt{15}}{3} by -\frac{80}{27} by multiplying \frac{-160+40\sqrt{15}}{3} by the reciprocal of -\frac{80}{27}.
x=\frac{-40\sqrt{15}-160}{-\frac{80}{27}\times 3}
Now solve the equation x=\frac{-\frac{160}{3}±\frac{40\sqrt{15}}{3}}{-\frac{80}{27}} when ± is minus. Subtract \frac{40\sqrt{15}}{3} from -\frac{160}{3}.
x=\frac{9\sqrt{15}}{2}+18
Divide \frac{-160-40\sqrt{15}}{3} by -\frac{80}{27} by multiplying \frac{-160-40\sqrt{15}}{3} by the reciprocal of -\frac{80}{27}.
x=-\frac{9\sqrt{15}}{2}+18 x=\frac{9\sqrt{15}}{2}+18
The equation is now solved.
-\frac{40}{27}x^{2}+\frac{160}{3}x-30=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.
-\frac{40}{27}x^{2}+\frac{160}{3}x-30-\left(-30\right)=-\left(-30\right)
Add 30 to both sides of the equation.
-\frac{40}{27}x^{2}+\frac{160}{3}x=-\left(-30\right)
Subtracting -30 from itself leaves 0.
-\frac{40}{27}x^{2}+\frac{160}{3}x=30
Subtract -30 from 0.
\frac{-\frac{40}{27}x^{2}+\frac{160}{3}x}{-\frac{40}{27}}=\frac{30}{-\frac{40}{27}}
Divide both sides of the equation by -\frac{40}{27}, which is the same as multiplying both sides by the reciprocal of the fraction.
x^{2}+\frac{\frac{160}{3}}{-\frac{40}{27}}x=\frac{30}{-\frac{40}{27}}
Dividing by -\frac{40}{27} undoes the multiplication by -\frac{40}{27}.
x^{2}-36x=\frac{30}{-\frac{40}{27}}
Divide \frac{160}{3} by -\frac{40}{27} by multiplying \frac{160}{3} by the reciprocal of -\frac{40}{27}.
x^{2}-36x=-\frac{81}{4}
Divide 30 by -\frac{40}{27} by multiplying 30 by the reciprocal of -\frac{40}{27}.
x^{2}-36x+\left(-18\right)^{2}=-\frac{81}{4}+\left(-18\right)^{2}
Divide -36, the coefficient of the x term, by 2 to get -18. Then add the square of -18 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-36x+324=-\frac{81}{4}+324
Square -18.
x^{2}-36x+324=\frac{1215}{4}
Add -\frac{81}{4} to 324.
\left(x-18\right)^{2}=\frac{1215}{4}
Factor x^{2}-36x+324. 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-18\right)^{2}}=\sqrt{\frac{1215}{4}}
Take the square root of both sides of the equation.
x-18=\frac{9\sqrt{15}}{2} x-18=-\frac{9\sqrt{15}}{2}
Simplify.
x=\frac{9\sqrt{15}}{2}+18 x=-\frac{9\sqrt{15}}{2}+18
Add 18 to both sides of the equation.