-16t^2+25t+150=0

Solution for -16t^2+25t+150=0 equation:

-16t^2+25t+150=0

a = -16; b = 25; c = +150;
Δ = b2-4ac
Δ = 252-4·(-16)·150
Δ = 10225
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:

$\sqrt{\Delta}=\sqrt{10225}=\sqrt{25*409}=\sqrt{25}*\sqrt{409}=5\sqrt{409}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5\sqrt{409}}{2*-16}=\frac{-25-5\sqrt{409}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5\sqrt{409}}{2*-16}=\frac{-25+5\sqrt{409}}{-32}
$