${t}^{2}+4t-32=0$
Write $4t$ as a difference${t}^{2}+8t-4t-32=0$
Factor out $t$ from the expression$t \times \left( t+8 \right)-4t-32=0$
Factor out $-4$ from the expression$t \times \left( t+8 \right)-4\left( t+8 \right)=0$
Factor out $t+8$ from the expression$\left( t+8 \right) \times \left( t-4 \right)=0$
When the product of factors equals $0$, at least one factor is $0$$\begin{array} { l }t+8=0,\\t-4=0\end{array}$
Solve the equation for $t$$\begin{array} { l }t=-8,\\t-4=0\end{array}$
Solve the equation for $t$$\begin{array} { l }t=-8,\\t=4\end{array}$
The equation has $2$ solutions$\begin{array} { l }t_1=-8,& t_2=4\end{array}$