Answer
$t=3+\sqrt{6}$ or $t=3-\sqrt{6}$
Work Step by Step
$ t^{2}+3=6t\qquad$...add $-6t$ to both sides.
$ t^{2}-6t+3=0\qquad$... solve with the Quadractic formula. $a=1,\ b=-6,\ c=3$
$ t=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-6,\ a$ for $1$ and $c$ for $3$.
$ t=\displaystyle \frac{-(-6)\pm\sqrt{(-6)^{2}-4\cdot(3)\cdot 1}}{2\cdot 1}\qquad$... simplify.
$t=\displaystyle \frac{6\pm\sqrt{36-12}}{2}$
$t=\displaystyle \frac{6\pm\sqrt{24}}{2}$
$t=\displaystyle \frac{6\pm\sqrt{4\cdot 6}}{2}$
$t=\displaystyle \frac{6\pm 2\sqrt{6}}{2}$
$ t=3\pm\sqrt{6}\qquad$... the symbol $\pm$ indicates two solutions.
$t=3+\sqrt{6}$ or $t=3-\sqrt{6}$
