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

Solution for -16t^2+10t+2=0 equation:

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

a = -16; b = 10; c = +2;
Δ = b2-4ac
Δ = 102-4·(-16)·2
Δ = 228
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:

$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{57}}{2*-16}=\frac{-10-2\sqrt{57}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{57}}{2*-16}=\frac{-10+2\sqrt{57}}{-32}
$