Answer
The solutions are $t=5+\sqrt{2}$ or $t=5-\sqrt{2}$
Work Step by Step
$ t^{2}-10t=-23\qquad$..add $25$ to both sides to complete the square ($\displaystyle \frac{1}{2}(-10)=-5$, and $(-5)^{2}=25$.)
$ t^{2}-10t+25=-23+25\qquad$...simplify by applying
the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms.
$(t-5)^{2}=2$
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-5=\pm\sqrt{2}\qquad$...add $5$ to each side.
$ t-5+5=\pm\sqrt{2}+5\qquad$...simplify.
$t=5\pm\sqrt{2}$
$t=5+\sqrt{2}$ or $t=5-\sqrt{2}$
