Solving $\frac{5}{t-3}-2=\frac{30}{t^2-9}$ I need help with solving this equation:
$$
\frac{5}{t-3}-2=\frac{30}{t^2-9}
$$
I tried to solve, but I always get false result. The result should be $-\frac{1}{2}$ but I always get $-\frac{1}{2}$ and $3$.
This is how I did it:
$$\begin{align}
\frac{5}{t-3}-2&=\frac{30}{t^2-9}\\
\frac{5-2\cdot(t-3)}{t-3}&=\frac{30}{t^2-9}\\
\frac{5-2t+6}{t-3}&=\frac{30}{t^2-9}\\
\frac{-2t+11}{t-3}&=\frac{30}{t^2-9}\\
\frac{-2t+11}{t-3}\cdot\left(t^2-9\right)&=30\\
\frac{(-2t+11)\cdot\left(t^2-9\right)}{t-3}&=30\\
\frac{(-2t+11)\cdot(t-3)\cdot(t+3)}{t-3}&=30\\
(-2t+11)\cdot(t+3)&=30\\
-2t^2-6t+11t+33&=30\\
-2t^2+5t+33&=30\\
-2t^2+5t+33-30&=0\\
-2t^2+5t+3&=0\\
\end{align}$$
Solving 
$$
-2t²+5t+3=0
$$
I get:
$$
-\frac{1}{2}\text{ and }3
$$
Am I doing something wrong?

You have cancelled out $t-3$ which is only allowed if $t-3\ne0\iff t\ne3$
