Solve for W
$\displaystyle
V=\sqrt{2Lw+w^2}
$
My answer was $\displaystyle
\Longrightarrow
w=V^2-2L
$
But apparently thats wrong according to my teacher.
Unfortunately you forgot to divide the Vē by w.
If you have a quadratic equation :
$\displaystyle ax^2+bx+c = 0$
Then this equation has the solutions:
$\displaystyle w = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
With your problem you have:
a = 1
b = -2L
c = -Vē
Plug in these terms into the formula to get the solutions.
(For confirmation only: $\displaystyle w = L\pm\sqrt{L^2+V^2}$
shouldn't it be a $\displaystyle -2L$ instead of $\displaystyle 2L$? According to the equation $\displaystyle w^2+2Lw-V^2=0$
$\displaystyle
w=\frac{-2L\pm\sqrt{4L^2+4v^2}}{2}
$
So the answer would be $\displaystyle
w = -L\pm\sqrt{L^2+V^2}
$