-16t^2+4t+10=0

Solution for -16t^2+4t+10=0 equation:

-16t^2+4t+10=0

a = -16; b = 4; c = +10;
Δ = b2-4ac
Δ = 42-4·(-16)·10
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{41}}{2*-16}=\frac{-4-4\sqrt{41}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{41}}{2*-16}=\frac{-4+4\sqrt{41}}{-32}
$