Answer
$t=-5+i$ or $t=-5-i$
Work Step by Step
$ t^{2}+10t+26=0\qquad$... solve with the Quadractic formula. $a=1,\ b=10,\ c=26$
$ t=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $10,\ a$ for $1$ and $c$ for $26$.
$ t=\displaystyle \frac{-10\pm\sqrt{(10)^{2}-4\cdot(26)\cdot 1}}{2\cdot 1}\qquad$... simplify.
$t=\displaystyle \frac{-10\pm\sqrt{100-104}}{2}$
$ t=\displaystyle \frac{-10\pm\sqrt{-4}}{2}\qquad$... write in terms of $i$. ($\sqrt{-1}=i$)
$t=\displaystyle \frac{-10\pm 2i}{2}$
$ t=-5\pm i\qquad$... the symbol $\pm$ indicates two solutions.
$t=-5+i$ or $t=-5-i$
