-16t^2+48t+32=0

Solution for -16t^2+48t+32=0 equation:

-16t^2+48t+32=0

a = -16; b = 48; c = +32;
Δ = b2-4ac
Δ = 482-4·(-16)·32
Δ = 4352
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:

$\sqrt{\Delta}=\sqrt{4352}=\sqrt{256*17}=\sqrt{256}*\sqrt{17}=16\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-16\sqrt{17}}{2*-16}=\frac{-48-16\sqrt{17}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+16\sqrt{17}}{2*-16}=\frac{-48+16\sqrt{17}}{-32}
$