-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 = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16}{2*-16}=\frac{32}{-32}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16}{2*-16}=\frac{64}{-32}
=-2
$