Solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ I want to solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ for $x$
I try this;
$5(2x+6)+2(x+3)=4(x+3)(2x+6)$
$12x+36 = 4(2x^2+12x+18)$
$8x^2+36x+36=0$
Where I would then go to the formula, however the answer says -1.5, what am I doing wrong.

The solutions of $8x^2+36x+36=0$ are $-1.5$ and $-3$, but the latter is not a solution of the original equation. This new, false solution comes when you multiply the equation by $(x+3)(2x+6)$, which is not the lcm of the denominators.
Note that $2x+6=2(x+3)$.
An example to show what's happening:
Take the equation
$$3x+5=-1$$
The solution is $x=-2$. If you multiply both sides of any equation by a non zero number, you get another equation, whose solutions are the same. But if you multiply both sides by $0$ you get $0=0$.
Now, if you multiply both sides by, say, $x+4$, we don't know if $x+4$ is zero or not. So we obtain another equation with two solutions:
$$3x^2+17x+20=-x-4$$
$$3x^3+18x+24=0$$
$$x=\left\lbrace\begin{align}&-2\\&-4\end{align}\right.$$
One of the solutions is the original one. The other is the value for which the factor $x+4$, that we multiplied by, vanishes.
