Answer
$t=3+i$ or $t=3-i$
Work Step by Step
$ t^{2}+10=6t\qquad$...add $-6t$ to both sides.
$ t^{2}-6t+10=0\qquad$... solve with the Quadractic formula. $a=1,\ b=-6,\ c=10$
$ t=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-6,\ a$ for $1$ and $c$ for $10$.
$ t=\displaystyle \frac{-(-6)\pm\sqrt{(-6)^{2}-4\cdot(10)\cdot 1}}{2\cdot 1}\qquad$... simplify.
$t=\displaystyle \frac{6\pm\sqrt{36-40}}{2}$
$ t=\displaystyle \frac{6\pm\sqrt{-4}}{2}\qquad$... write in terms of $i$. ($\sqrt{-1}=i$)
$t=\displaystyle \frac{6\pm 2i}{2}$
$ t=3\pm i\qquad$... the symbol $\pm$ indicates two solutions.
$t=3+i$ or $t=3-i$
