Answer
$\color{blue}{-1, 4}$
Work Step by Step
Multiply $t-3$ to both sides of the equation to obtain:
$2t(t-3) = \dfrac{8}{t-3} \cdot (t-3)
\\2t(t)-2t(3)=8
\\2t^2-6t=8
\\2t^2-6t-8=0$
Factor out $2$ to obtain:
$2(t^2-3t-4)=0$
Divide $2$ to both sides to obtain:
$t^2-3t-4=0$
Factor the trinomial to obtain:
$(t-4)(t+1)=0$
Equate each factor to zero, then solve each equation to obtain:
\begin{array}{ccc}
&t-4=0 &\text{or} &t+1=0
\\&t=4 &\text{or} &t=-1
\end{array}
Cheking:
If t=4:
$2(4)=\dfrac{8}{4-3}?
\\8=\dfrac{8}{1}?
\\8=8$
If $t=-1$:
$2(-1)=\dfrac{8}{-1-3}?
\\-2 =\dfrac{8}{-4}?
\\-2=-2$
Thus, the solutions to the given equation are: $\color{blue}{-1, 4}$.
