#### Answer

$t=\pm 3$ or $t=\pm\sqrt{2}$

#### Work Step by Step

$ t^{4}-11t^{2}+18\qquad$...substitute $t^{2}$ for $u$
$ u^{2}-11u+18=0\qquad$... solve with the Quadractic formula. $a=1,\ b=-11,\ c=18$
$ u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-7,\ a$ for $1$ and $c$ for $18$.
$ u=\displaystyle \frac{-(-11)\pm\sqrt{(-11)^{2}-4\cdot(18)\cdot 1}}{2\cdot 1}\qquad$... simplify.
$u=\displaystyle \frac{11\pm\sqrt{121-72}}{2}$
$u=\displaystyle \frac{11\pm\sqrt{49}}{2}$
$ u=\displaystyle \frac{11\pm 7}{2}\qquad$... the symbol $\pm$ indicates two solutions.
$u=\displaystyle \frac{11+7}{2}=\frac{18}{2}=9$ or $u=\displaystyle \frac{11-7}{2}=\frac{4}{2}=2$
Bring back $t^{2}=u$.
$t^{2}=9$ or $t^{2}=2$
$t=\pm 3$ or $t=\pm\sqrt{2}$