-16t^2+48t+5.5=0

Solution for -16t^2+48t+5.5=0 equation:

-16t^2+48t+5.5=0

a = -16; b = 48; c = +5.5;
Δ = b2-4ac
Δ = 482-4·(-16)·5.5
Δ = 2656
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2656}=\sqrt{16*166}=\sqrt{16}*\sqrt{166}=4\sqrt{166}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-4\sqrt{166}}{2*-16}=\frac{-48-4\sqrt{166}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+4\sqrt{166}}{2*-16}=\frac{-48+4\sqrt{166}}{-32}
$