-16t^2+150t+10=0

Solution for -16t^2+150t+10=0 equation:

-16t^2+150t+10=0

a = -16; b = 150; c = +10;
Δ = b2-4ac
Δ = 1502-4·(-16)·10
Δ = 23140
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:

$\sqrt{\Delta}=\sqrt{23140}=\sqrt{4*5785}=\sqrt{4}*\sqrt{5785}=2\sqrt{5785}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(150)-2\sqrt{5785}}{2*-16}=\frac{-150-2\sqrt{5785}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(150)+2\sqrt{5785}}{2*-16}=\frac{-150+2\sqrt{5785}}{-32}
$