Is there an easier way to solve this integral problem? I have the following integral:
$$F(t) = \int_0^b (\sqrt{2 + t} - 2) dt$$
I need to find a positive value of $b$ such that $F(t) = 0$.
Going through the integration steps and using the Fundamental Theorem of Calculus, I end up with
\begin{align}
0 & = (\frac{2}{3}\sqrt{(2 + t)}^3 - 2t) |_0^b  \\
& = (\frac{2}{3}\sqrt{(2 + b)}^3 - 2b) - (\frac{2}{3}\sqrt{2}^3) \\
\Rightarrow \frac{4\sqrt{2}}{3} & = \frac{2}{3}\sqrt{(2 + b)}^3 - 2b
\end{align}
But isolating $b$ here is very complicated, and WolframAlpha gives a very ugly solution. It's been a while since I took calculus, but I feel like I'm missing something obvious.

If you  change variable $b=x^2-2$, the equation becomes $$ \frac{2 x^3}{3}-2 x^2+4\big(1-\frac{ \sqrt{2}}{3}\big)=0$$ which can be solved with radicals using Cardano method.
But, there is one "obvious" root $x_1=\sqrt 2$ (remember that this corresponds to $b=0$ which is a trivial solution of the problem). So, what is left is a quadratic equation $$\frac{2 x^2}{3}+\left(\frac{2 \sqrt{2}}{3}-2\right) x-\big(2 \sqrt{2}-\frac{4}{3}\big)=0$$ the roots of which being $$x_2=\frac{1}{2} \left(3-\sqrt{2}-\sqrt{3+6 \sqrt{2}}\right)$$ $$x_3=\frac{1}{2} \left(3-\sqrt{2}+\sqrt{3+6 \sqrt{2}}\right)$$ The root $x_2$ must be discarded since it is negative and the only one left is then $x_3$; so, after simplifications, $$b=x_3^2-2=\frac{1}{2} \left(3+\sqrt{48 \sqrt{2}-39}\right)\approx 4.18711$$ which is quite close to  Shailesh's answer.
