-16t2-10t+6=0

Solution for -16t2-10t+6=0 equation:

-16t^2-10t+6=0

a = -16; b = -10; c = +6;
Δ = b2-4ac
Δ = -102-4·(-16)·6
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-22}{2*-16}=\frac{-12}{-32}
=3/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+22}{2*-16}=\frac{32}{-32}
=-1
$