Answer
The solutions are $t=-4+\sqrt{19}$ or $t=-4-\sqrt{19}$.
Work Step by Step
$ t^{2}+8t-3=0\qquad$..add $3$ to both sides so we can complete the square on the left side.
$ t^{2}+8t=3\qquad$..add $16$ to both sides to complete the square ($\displaystyle \frac{1}{2}(8)=4$, and $(4)^{2}=16$.)
$ t^{2}+8t+16=3+16\qquad$...simplify by applying
the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms.
$(t+4)^{2}=19$
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+4=\pm\sqrt{19}\qquad$...add $-4$ to each side.
$ t+4-4=\pm\sqrt{19}-4\qquad$...simplify.
$t=-4\pm\sqrt{19}$
$t=-4+\sqrt{19}$ or $t=-4-\sqrt{19}$
