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3t^{2}-17t+10=0
Substitute t for x^{2}.
t=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 3\times 10}}{2\times 3}
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 3 for a, -17 for b, and 10 for c in the quadratic formula.
t=\frac{17±13}{6}
Do the calculations.
t=5 t=\frac{2}{3}
Solve the equation t=\frac{17±13}{6} when ± is plus and when ± is minus.
x=\sqrt{5} x=-\sqrt{5} x=\frac{\sqrt{6}}{3} x=-\frac{\sqrt{6}}{3}
Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for each t.