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t^{2}+5t-1=0
Substitute t for x^{5}.
t=\frac{-5±\sqrt{5^{2}-4\times 1\left(-1\right)}}{2}
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 1 for a, 5 for b, and -1 for c in the quadratic formula.
t=\frac{-5±\sqrt{29}}{2}
Do the calculations.
t=\frac{\sqrt{29}-5}{2} t=\frac{-\sqrt{29}-5}{2}
Solve the equation t=\frac{-5±\sqrt{29}}{2} when ± is plus and when ± is minus.
x=\sqrt[5]{\frac{\sqrt{29}-5}{2}} x=\sqrt[5]{\frac{-\sqrt{29}-5}{2}}
Since x=t^{5}, the solutions are obtained by evaluating x=\sqrt[5]{t} for each t.