Answer
$t=7.25$, $t=-8 $
Work Step by Step
Given \begin{equation}
4 t^2+3 t=232.
\end{equation}
This is a quadratic equation because of the square term. Solve it using the quadratic formula.
\begin{equation}
\begin{aligned}
4 t^2+3 t&=232 \\
4 t^2+3 t-232&= 0\\
ax^2+bx+c&=0\\
a & =4,\ b=3,\ c=-232 \\
t & =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
t & =\frac{-(3) \pm \sqrt{(3)^2-4 \cdot 4 \cdot(-232)}}{2 \cdot 4} \\
& =\frac{-3 \pm 61}{8} \\
\Longrightarrow t& =\frac{-3+ 61}{8}\\
&= 7.25\\
t& =\frac{-3 -61}{8} \\
&= -8.
\end{aligned}
\end{equation} The solution is $t=7.25$, $t=-8 $.
