Answer
The solutions are $t=-3+\sqrt{14}$ or $t=-3-\sqrt{14}$.
Work Step by Step
$ t^{2}+6t-5=0\qquad$..add $5$ to both sides so we can complete the square on the left side.
$ t^{2}+6t=5\qquad$..add $9$ to both sides to complete the square ($\displaystyle \frac{1}{2}(6)=3$, and $(3)^{2}=9$.)
$ t^{2}+6t+9=5+9\qquad$...simplify by applying
the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms.
$(t+3)^{2}=14$
According to the general principle of square roots:
For any real number $k$ and any algebraic expression $x$ :
$\text{If }x^{2}=k,\text{ then }x=\sqrt{k}\text{ or }x=-\sqrt{k}$.
$ t+3=\pm\sqrt{14}\qquad$...add $-3$ to each side.
$ t+3-3=\pm\sqrt{14}-3\qquad$...simplify.
$t=-3\pm\sqrt{14}$
$t=-3+\sqrt{14}$ or $t=-3-\sqrt{14}$
