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9t^{2}-10t+1=0
Substitute t for m^{2}.
t=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 9\times 1}}{2\times 9}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 9 for a, -10 for b, and 1 for c in the quadratic formula.
t=\frac{10±8}{18}
Do the calculations.
t=1 t=\frac{1}{9}
Solve the equation t=\frac{10±8}{18} when ± is plus and when ± is minus.
m=1 m=-1 m=\frac{1}{3} m=-\frac{1}{3}
Since m=t^{2}, the solutions are obtained by evaluating m=±\sqrt{t} for each t.