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\frac{5}{27}\times \frac{1}{10^{21}}=t^{2}
Reduce the fraction \frac{40}{216} to lowest terms by extracting and canceling out 8.
\frac{5}{27}\times \frac{1}{1000000000000000000000}=t^{2}
Calculate 10 to the power of 21 and get 1000000000000000000000.
\frac{1}{5400000000000000000000}=t^{2}
Multiply \frac{5}{27} and \frac{1}{1000000000000000000000} to get \frac{1}{5400000000000000000000}.
t^{2}=\frac{1}{5400000000000000000000}
Swap sides so that all variable terms are on the left hand side.
t=\frac{\sqrt{6}}{180000000000} t=-\frac{\sqrt{6}}{180000000000}
Take the square root of both sides of the equation.
\frac{5}{27}\times \frac{1}{10^{21}}=t^{2}
Reduce the fraction \frac{40}{216} to lowest terms by extracting and canceling out 8.
\frac{5}{27}\times \frac{1}{1000000000000000000000}=t^{2}
Calculate 10 to the power of 21 and get 1000000000000000000000.
\frac{1}{5400000000000000000000}=t^{2}
Multiply \frac{5}{27} and \frac{1}{1000000000000000000000} to get \frac{1}{5400000000000000000000}.
t^{2}=\frac{1}{5400000000000000000000}
Swap sides so that all variable terms are on the left hand side.
t^{2}-\frac{1}{5400000000000000000000}=0
Subtract \frac{1}{5400000000000000000000} from both sides.
t=\frac{0±\sqrt{0^{2}-4\left(-\frac{1}{5400000000000000000000}\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{1}{5400000000000000000000} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{0±\sqrt{-4\left(-\frac{1}{5400000000000000000000}\right)}}{2}
Square 0.
t=\frac{0±\sqrt{\frac{1}{1350000000000000000000}}}{2}
Multiply -4 times -\frac{1}{5400000000000000000000}.
t=\frac{0±\frac{\sqrt{6}}{90000000000}}{2}
Take the square root of \frac{1}{1350000000000000000000}.
t=\frac{\sqrt{6}}{180000000000}
Now solve the equation t=\frac{0±\frac{\sqrt{6}}{90000000000}}{2} when ± is plus.
t=-\frac{\sqrt{6}}{180000000000}
Now solve the equation t=\frac{0±\frac{\sqrt{6}}{90000000000}}{2} when ± is minus.
t=\frac{\sqrt{6}}{180000000000} t=-\frac{\sqrt{6}}{180000000000}
The equation is now solved.