Answer
$t=2, t=-12$
Work Step by Step
We start by completing the square in line 1 of the equation by dividing the coefficient of the $10t$ term by $2$ and squaring it and adding the result on both sides.
$$
\begin{aligned}
t^2+10t & = 24 \\
t^2+ 10t+\left(\frac{10}{2}\right)^2 & =24+\left(\frac{10}{2}\right)^2 \\
t^2+10t+5^2 & =24+5^2 \\
(t+5)^2 & =49.
\end{aligned}
$$ Take the square root:
$$
\begin{array}{rl}
(t+5)^2 & =49 \\
t+5 & = \pm \sqrt{49} \\
t+5 & = \pm 7 \\
t+5=7 &\ \textbf{or}\ \ \ t+5=-7\\
t=7-5 = 2& \ \textbf{or}\ \ \ t=-7-5= -12.
\end{array}
$$ Check $$\begin{aligned}
2^2+10(2) & \stackrel{?}{=} 24\\
4+20& \stackrel{?}{=} 24\\
24& = 24\checkmark
\end{aligned}$$ $$\begin{aligned}
(-12)^2+10(-12) & \stackrel{?}{=} 24\\
144-120& \stackrel{?}{=} 24\\
24& = 24\checkmark.
\end{aligned}$$ The solution is: $t=2, t=-12$.
