-16t^2+122t+10=0

Solution for -16t^2+122t+10=0 equation:

-16t^2+122t+10=0

a = -16; b = 122; c = +10;
Δ = b2-4ac
Δ = 1222-4·(-16)·10
Δ = 15524
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:

$\sqrt{\Delta}=\sqrt{15524}=\sqrt{4*3881}=\sqrt{4}*\sqrt{3881}=2\sqrt{3881}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(122)-2\sqrt{3881}}{2*-16}=\frac{-122-2\sqrt{3881}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(122)+2\sqrt{3881}}{2*-16}=\frac{-122+2\sqrt{3881}}{-32}
$