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9t^{2}-46t+5=0
Substitute t for n^{2}.
t=\frac{-\left(-46\right)±\sqrt{\left(-46\right)^{2}-4\times 9\times 5}}{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, -46 for b, and 5 for c in the quadratic formula.
t=\frac{46±44}{18}
Do the calculations.
t=5 t=\frac{1}{9}
Solve the equation t=\frac{46±44}{18} when ± is plus and when ± is minus.
n=\sqrt{5} n=-\sqrt{5} n=\frac{1}{3} n=-\frac{1}{3}
Since n=t^{2}, the solutions are obtained by evaluating n=±\sqrt{t} for each t.