Answer
$x=\pm 2$ or $x=\pm\sqrt{3}$
Work Step by Step
$ t^{4}-7t^{2}+12=0\qquad$...substitute $t^{2}$ for $u$
$ u^{2}-7u+12=0\qquad$... solve with the Quadractic formula. $a=1,\ b=-7,\ c=12$
$ u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-7,\ a$ for $1$ and $c$ for $12$.
$ u=\displaystyle \frac{-(-7)\pm\sqrt{(-7)^{2}-4\cdot(12)\cdot 1}}{2\cdot 1}\qquad$... simplify.
$u=\displaystyle \frac{7\pm\sqrt{49-48}}{2}$
$u=\displaystyle \frac{7\pm\sqrt{1}}{2}$
$ u=\displaystyle \frac{7\pm 1}{2}\qquad$... the symbol $\pm$ indicates two solutions.
$u=\displaystyle \frac{7+1}{2}=\frac{8}{2}=4$ or $u=\displaystyle \frac{7-1}{2}=\frac{6}{2}=3$
Bring back $x$.
$x^{2}=4$ or $x^{2}=3$
$x=\pm 2$ or $x=\pm\sqrt{3}$
