-16t^2+72t+40=0

Solution for -16t^2+72t+40=0 equation:

-16t^2+72t+40=0

a = -16; b = 72; c = +40;
Δ = b2-4ac
Δ = 722-4·(-16)·40
Δ = 7744
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7744}=88$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-88}{2*-16}=\frac{-160}{-32}
=+5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+88}{2*-16}=\frac{16}{-32}
=-1/2
$