-32=48t-16t^2

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

-32=48t-16t^2

We move all terms to the left:

-32-(48t-16t^2)=0

We get rid of parentheses

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}
$