-16t2+48t+64=0

Solution for -16t2+48t+64=0 equation:

-16t^2+48t+64=0

a = -16; b = 48; c = +64;
Δ = b2-4ac
Δ = 482-4·(-16)·64
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-80}{2*-16}=\frac{-128}{-32}
=+4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+80}{2*-16}=\frac{32}{-32}
=-1
$