Solve for z
z=1
z=\sqrt[3]{2}\approx 1.25992105
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t^{2}-3t+2=0
Substitute t for z^{3}.
t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 2}}{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, -3 for b, and 2 for c in the quadratic formula.
t=\frac{3±1}{2}
Do the calculations.
t=2 t=1
Solve the equation t=\frac{3±1}{2} when ± is plus and when ± is minus.
z=\sqrt[3]{2} z=1
Since z=t^{3}, the solutions are obtained by evaluating z=\sqrt[3]{t} for each t.
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