-16t2+20t+2=0

Solution for -16t2+20t+2=0 equation:

-16t^2+20t+2=0

a = -16; b = 20; c = +2;
Δ = b2-4ac
Δ = 202-4·(-16)·2
Δ = 528
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{33}}{2*-16}=\frac{-20-4\sqrt{33}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{33}}{2*-16}=\frac{-20+4\sqrt{33}}{-32}
$