96t^2-40t-16=0

Solution for 96t^2-40t-16=0 equation:

96t^2-40t-16=0

a = 96; b = -40; c = -16;
Δ = b2-4ac
Δ = -402-4·96·(-16)
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{-(-40)-88}{2*96}=\frac{-48}{192}
=-1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+88}{2*96}=\frac{128}{192}
=2/3
$