Answer
$t=-2+\sqrt{5}$ or $t=-2-\sqrt{5}$
Work Step by Step
$ t^{2}+4t=1\qquad$...square half the coefficient of $x$ and add it to both sides to complete the square
$(\displaystyle \frac{4}{2})=2,\ (2)^{2}=4\qquad$...add $4$ to both sides.
$ t^{2}+4t+4=5\\qquad$...write $t^{2}+4t+4$ as a binomial squared.
$(t+2)^{2}=5\qquad$...use the 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+2=\pm\sqrt{5}\qquad$...add $-2$ to both sides.
$ t=-2\pm\sqrt{5}\qquad$... the symbol $\pm$ indicates two solutions.
$t=-2+\sqrt{5}$ or $t=-2-\sqrt{5}$
