-16t2+32t+48=0

Solution for -16t2+32t+48=0 equation:

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

a = -16; b = 32; c = +48;
Δ = b2-4ac
Δ = 322-4·(-16)·48
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-64}{2*-16}=\frac{-96}{-32}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+64}{2*-16}=\frac{32}{-32}
=-1
$