-16t^2+40t+144=0

Solution for -16t^2+40t+144=0 equation:

-16t^2+40t+144=0

a = -16; b = 40; c = +144;
Δ = b2-4ac
Δ = 402-4·(-16)·144
Δ = 10816
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10816}=104$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-104}{2*-16}=\frac{-144}{-32}
=4+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+104}{2*-16}=\frac{64}{-32}
=-2
$