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