16t^2-24t-16=0

Solution for 16t^2-24t-16=0 equation:

16t^2-24t-16=0

a = 16; b = -24; c = -16;
Δ = b2-4ac
Δ = -242-4·16·(-16)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-40}{2*16}=\frac{-16}{32}
=-1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+40}{2*16}=\frac{64}{32}
=2
$