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

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

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

a = -16; b = 10; c = +150;
Δ = b2-4ac
Δ = 102-4·(-16)·150
Δ = 9700
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:

$\sqrt{\Delta}=\sqrt{9700}=\sqrt{100*97}=\sqrt{100}*\sqrt{97}=10\sqrt{97}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{97}}{2*-16}=\frac{-10-10\sqrt{97}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{97}}{2*-16}=\frac{-10+10\sqrt{97}}{-32}
$