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