Answer
$-1-\sqrt 6$;$-1+\sqrt 6$
Work Step by Step
First transfer the constant to the right of the equal sign proceed with completing the square in line $1$ of the equation by dividing the coefficient of the $2t$ term by 2 and squaring it and adding the result on both sides. $$
\begin{aligned}
h^2+2t -5& = 0 \\
h^2+2t & = 5 \\
h^2+2t+\left(\frac{2}{2}\right)^2 & =5+\left(\frac{2}{2}\right)^2 \\
h^2+2t+1^2 & =5+1^2 \\
(h+1)^2 & = 6.
\end{aligned}
$$ Take the square root:
$$
\begin{array}{rl}
(h+1)^2 & = 6 \\
h+1 & = \pm \sqrt{6} \\
h+1=\sqrt{6} &\ \textbf{or}\ \ \ h+1=-\sqrt{6}\\
h=-1+\sqrt{6} & \ \textbf{or}\ \ \ h= -1-\sqrt{6}.
\end{array}
$$ Check: $$\begin{aligned}
(-1+\sqrt{6})^2+2(-1+\sqrt{6}) -5&\stackrel{?}{=} 0 \\
1-2\sqrt 6+6-2+2\sqrt 6-5& \stackrel{?}{=} 0\\
0&=0\checkmark\\
(-1-\sqrt{6})^2+2(-1-\sqrt{6}) -5&\stackrel{?}{=} 0 \\
1+2\sqrt 6+6-2-2\sqrt 6-5& \stackrel{?}{=} 0\\
0&=0\checkmark.
\end{aligned}$$ The solution is:
$$h=-1-\sqrt 6, \quad \ h=-1+\sqrt 6.$$
