What is the mistake in my solution to find roots of a parabola? The parabola is,
$\frac{1}{2}g{{t}_{t}}^{2}+u{t}_{t}-2H=0$
By the way, $u=\sqrt{\frac{3gH}{2}}$. I have found the discriminant as below,
$\Delta=b^2-4ac$
$\Delta=u^2-4(\frac{1}{2}g)(-2H)$
$\Delta=u^2+4gH$
$u^2=\frac{3gH}{2} \Rightarrow \Delta=\frac{3gH}{2}+4gH=\frac{11gH}{2}$
Let's find roots,
${{t}_{t}}_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}$
${{t}_{t}}_{1,2}=\frac{-u\pm\sqrt{\frac{11gH}{2}}}{g}$
The variable ${{t}_{t}}$ is a variable of time. Time can not be negative, so I use $+$ sign.
${{t}_{t}}=\frac{-u+\sqrt{\frac{11gH}{2}}}{g}$
But, when I put this in the parabola, I can't get $0$ as result. Is there any mistake?
important: $g$ and $H$ are constants. So, $u$ as well a constant.

Both solutions produce $0$ in $f$.
\begin{align*}
f&\left( \frac{-u + \sqrt{\frac{11gH}{2}}}{g} \right)  \\
    &= \frac{g}{2} \left( \frac{-u + \sqrt{\frac{11gH}{2}}}{g} \right)^2 + u \left( \frac{-u + \sqrt{\frac{11gH}{2}}}{g} \right) - 2H  \\
    &= \frac{g}{2} \left( \frac{ \left( -u + \sqrt{\frac{11gH}{2}} \right)^2}{g^2} \right) + u \left( \frac{-u + \sqrt{\frac{11gH}{2}}}{g} \right) - 2H  \\
    &= \frac{1}{2} \cdot \frac{\left( -u + \sqrt{\frac{11gH}{2}} \right)^2}{g} + u \left( \frac{-u + \sqrt{\frac{11gH}{2}}}{g} \right) - 2H \cdot \frac{g}{g}  \\
    &= \frac{1}{2} \cdot \frac{u^2 -2u\sqrt{\frac{11gH}{2}} + \frac{11gH}{2}}{g} +  \frac{-u^2 + u \sqrt{\frac{11gH}{2}}}{g} - \frac{2gH}{g}  \\
    &= \frac{\frac{1}{2} u^2 -u\sqrt{\frac{11gH}{2}} + \frac{11gH}{4}}{g} +  \frac{-u^2 + u \sqrt{\frac{11gH}{2}}}{g} - \frac{2gH}{g}  \\
    &= \frac{\frac{1}{2} u^2 -u\sqrt{\frac{11gH}{2}} + \frac{11gH}{4} -u^2 + u \sqrt{\frac{11gH}{2}} -2gH}{g}  \\
    &= \frac{\frac{-1}{2} u^2 + \frac{11gH}{4} - 2gH\cdot\frac{4}{4}}{g}  \\
    &= \frac{\frac{-1}{2} \frac{3gH}{2} + \frac{11gH}{4} - \frac{8gH}{4}}{g}  \\
    &= \frac{\frac{-3gH}{4} + \frac{11gH}{4} - \frac{8gH}{4}}{g}  \\
    &= \frac{0}{g}  \\
    &= 0  \text{.}
\end{align*}
Similarly for the other root.
