-16t^2+54t+40=0

Solution for -16t^2+54t+40=0 equation:

-16t^2+54t+40=0

a = -16; b = 54; c = +40;
Δ = b2-4ac
Δ = 542-4·(-16)·40
Δ = 5476
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5476}=74$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-74}{2*-16}=\frac{-128}{-32}
=+4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+74}{2*-16}=\frac{20}{-32}
=-5/8
$