Answer
$\dfrac{-11+\sqrt{137}}{2}$; $\dfrac{-11-\sqrt{137}}{2}$
Work Step by Step
We start by completing the square in line $1$ of the equation by dividing the coefficient of the $11t$ term by $2$ and squaring it and adding the result on both sides.
$$
\begin{aligned}
t^2+11t & = 4 \\
t^2+11t & = 4 \\
t^2+11t+\left(\frac{11}{2}\right)^2 & =4+\left(\frac{11}{2}\right)^2 \\
t^2+11t+\frac{121}{4} & =4+\frac{121}{4} \\
\left(t+\frac{11}{2}\right)^2 & =\frac{137}{4}.
\end{aligned}
$$ Take the square root: $$
\begin{array}{rl}
t+\dfrac{11}{2} & = \pm \dfrac{\sqrt{137}}{2} \\\\
t+\dfrac{11}{2}=\dfrac{\sqrt{137}}{2} &\ \textbf{or}\ \ \ t+\dfrac{11}{2}=-\dfrac{\sqrt{137}}{2}\\
t=\dfrac{-11+\sqrt{137}}{2} & \ \textbf{or}\ \ \ t=\dfrac{-11-\sqrt{137}}{2}.
\end{array}
$$ Check $$\begin{aligned}
\left(\dfrac{-11+\sqrt{137}}{2}\right)^2+11\left(\dfrac{-11+\sqrt{137}}{2}\right) & \stackrel{?}{=} 4 \\ \dfrac{121-22\sqrt{137}+137-242+22\sqrt{137}}{4}&\stackrel{?}{=}4\\
4& = 4\checkmark\\
\left(\dfrac{-11-\sqrt{137}}{2}\right)^2+11\left(\dfrac{-11-\sqrt{137}}{2}\right) & \stackrel{?}{=} 4 \\ \dfrac{121+22\sqrt{137}+137-242-22\sqrt{137}}{4}&\stackrel{?}{=}4\\
4& = 4\checkmark.
\end{aligned}$$ The solution is: $$t=\dfrac{-11+\sqrt{137}}{2} \quad \ \textbf{or}\ \ \ t=\dfrac{-11-\sqrt{137}}{2}.$$
