Answer
There are no real solutions.
Work Step by Step
$ t^{\frac{1}{2}}+3t^{\frac{1}{4}}+2=0\qquad$...substitute $t^{\frac{1}{4}}$ for $u$ so that $u^{2}=t^{\frac{1}{2}}$
Note that $t^{\frac{1}{4}}$ is the fourth root of $t$, a positive number.
$ u^{2}+3u+2=0\qquad$... solve with the Quadratic formula.
$u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$u=\displaystyle \frac{-3\pm\sqrt{9-8}}{2}$
$u=\displaystyle \frac{-3\pm\sqrt{1}}{2}$
$u=\displaystyle \frac{-3\pm 1}{2}$
$u=\displaystyle \frac{-3+1}{2}=\frac{-2}{2}=-1$ or $u=\displaystyle \frac{-3-1}{2}=\frac{-4}{2}=-2$
Bring back $t^{\frac{1}{4}}=u$.
$t^{\frac{1}{4}}=-1$ or $t^{\frac{1}{4}}=-2$
We discard both solutions because the fourth root of $t$ is nonnegative.
There are no real solutions.
