Answer
$t=-\dfrac{1}{2}$ and $t=-6$
Work Step by Step
$\dfrac{2t+3}{t-1}-\dfrac{2}{t+3}=\dfrac{5-6t}{t^{2}+2t-3}$
Factor the denominator of the third fraction:
$\dfrac{2t+3}{t-1}-\dfrac{2}{t+3}=\dfrac{5-6t}{(t+3)(t-1)}$
Multiply the whole equation by $(t+3)(t-1)$:
$(t+3)(t-1)\Big[\dfrac{2t+3}{t-1}-\dfrac{2}{t+3}=\dfrac{5-6t}{(t+3)(t-1)}\Big]$
$(2t+3)(t+3)-2(t-1)=5-6t$
$2t^{2}+9t+9-2t+2=5-6t$
Take all terms to the left side of the equation and simplify it by combining like terms:
$2t^{2}+9t+9-2t+2-5+6t=0$
$2t^{2}+13t+6=0$
Solve this equation by factoring:
$(2t+1)(t+6)=0$
The two solutions are:
$t=-\dfrac{1}{2}$ and $t=-6$
The original equation is not undefined for neither of these values found. The answer is:
$t=-\dfrac{1}{2}$ and $t=-6$
