-16t^2+150t+144=0

Solution for -16t^2+150t+144=0 equation:

-16t^2+150t+144=0

a = -16; b = 150; c = +144;
Δ = b2-4ac
Δ = 1502-4·(-16)·144
Δ = 31716
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:

$\sqrt{\Delta}=\sqrt{31716}=\sqrt{36*881}=\sqrt{36}*\sqrt{881}=6\sqrt{881}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(150)-6\sqrt{881}}{2*-16}=\frac{-150-6\sqrt{881}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(150)+6\sqrt{881}}{2*-16}=\frac{-150+6\sqrt{881}}{-32}
$