Question: I am currently stuck on a problem that should be easy enough... In an isosceles triangle the two equal sides are $5m$ and the area is $12m^2$. How can I find out the length of the third side? 

My approach so far is this:

$24 = b·h$

$5^2 = (\frac{b}{2})^2 · h^2$

When you do the substitution you always get the nasty $x^4$ type equation. Any one that can help me here?

Answer: 
I think you want to use $\displaystyle 5^2 = \left(\frac{b}{2}\right)^2 + h^2\tag{1},$   

assuming you mean to be using the Pythagorean Theorem - you need to sum the terms on the right of $(1)$, in contrast to what you've written in your question: $ 5^2 = \left(\frac{b}{2}\right)^2 \cdot h^2$. 

$24 = b\cdot h \implies h = \dfrac{24}{b} \tag{2}$

So from $(1)$ we get $\quad 25 = \dfrac{b^2}{4} + h^2\tag{3}.$

Substitute your value for $h$ (found from $(2)$) into $(3)$, and solve for $b$. 

$$25 = \dfrac{b^2}{4} + \left(\dfrac{24}{b}\right)^2 \implies 100b^2 = b^4 + 4\cdot24^2\tag{3}\implies b^4 - 100 b^2 + 2304 = 0$$ $$\iff (b^2 - 36)(b^2 - 64) = 0\tag{4}$$

Both factors in $(4)$ are a "difference of squares": so there will be four solutions to $(4)$, two of which are negative, so you need to throw those out (can't have negative length!).   

That leaves you with two possible solutions for the base: $b\_1 = x\_1,$ or $b\_2=x\_2$, and respectively, when $b\_1 = x\_1, \;h\_1=\dfrac{x\_2}{2}$, or when $b\_2 = x\_2, \; h\_2 = \dfrac{ x\_1}{2}$.
It is convenient to let the unknown third side be $2b$. Let the height be $h$. The area condition yields $bh=12$. By the Pythagorean Theorem, $b^2+h^2=25$. 

Now instead of using a quadratic equation, we use a technique that goes back to Neo-Babylonian times. Note that
$$(b+h)^2=b^2+h^2+2bh=49.$$
Similarly, $(b-h)^2=b^2+h^2-2bh=1$. 

Thus $b+h=7$ and $b-h=\pm 1$. Add: We get $2b=8$ or $2b=6$.