-16t^2+150t+20=0

Solution for -16t^2+150t+20=0 equation:

-16t^2+150t+20=0

a = -16; b = 150; c = +20;
Δ = b2-4ac
Δ = 1502-4·(-16)·20
Δ = 23780
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:

$\sqrt{\Delta}=\sqrt{23780}=\sqrt{4*5945}=\sqrt{4}*\sqrt{5945}=2\sqrt{5945}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(150)-2\sqrt{5945}}{2*-16}=\frac{-150-2\sqrt{5945}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(150)+2\sqrt{5945}}{2*-16}=\frac{-150+2\sqrt{5945}}{-32}
$