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2t^{2}-4t+6=0
Substitute t for x^{2}.
t=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\times 6}}{2\times 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 2 for a, -4 for b, and 6 for c in the quadratic formula.
t=\frac{4±\sqrt{-32}}{4}
Do the calculations.
t\in \emptyset
Since the square root of a negative number is not defined in the real field, there are no solutions.
x\in \emptyset
Since t=x^{2}, the original equation does not have any solutions.