Hello, TriKri!
I too got a fourthdegree equation, too.
But there is a way around it . . .
[quote]A 10m ladder is leaned against a box (1 x 1 x 1 m) placed against a wall.
But the ground is so slippery that the ladder falls over the box and leans against the wall.
How high up on the wall does the ladder reach? Code:

*
* y
10 *    *
*  1 
* 1 1
*  
*        *    *
x 1
From the large right triangle: .$\displaystyle (y+1)^2 + (1+x)^2 \:=\:10^2$ .[1]
The two smaller right triangles are similar.
. . So we have: .$\displaystyle \frac{1}{x} \:=\:\frac{y}{1} \quad\Rightarrow\quad x \:=\:\frac{1}{y}$ .[2]
Substitute [2] into [1]: .$\displaystyle (y+1)^2 + \left(1 + \frac{1}{y}\right)^2 \:=\:100$
and we have: .$\displaystyle y^2 + 2y + 1 + 1 + \frac{2}{y} + \frac{1}{y^2} \;=\;100$
. . $\displaystyle \left(y^2 + 2 + \frac{1}{y^2}\right) + \left(2y + \frac{2}{y}\right)  100 \:=\:0 $
. . $\displaystyle \left(y + \frac{1}{y}\right)^2 + 2\left(y + \frac{1}{y}\right)  100 \:=\:0$
Let $\displaystyle u \:=\:y+\frac{1}{y}$
. . and we have the quadratic: .$\displaystyle u^2 + 2u  100 \:=\:100$
Use the Quadratic Formula to solve for $\displaystyle u$,
. . then backsubstitute and solve for $\displaystyle y.$