-16t2+150t-3=0

Solution for -16t2+150t-3=0 equation:

-16t^2+150t-3=0

a = -16; b = 150; c = -3;
Δ = b2-4ac
Δ = 1502-4·(-16)·(-3)
Δ = 22308
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:

$\sqrt{\Delta}=\sqrt{22308}=\sqrt{676*33}=\sqrt{676}*\sqrt{33}=26\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(150)-26\sqrt{33}}{2*-16}=\frac{-150-26\sqrt{33}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(150)+26\sqrt{33}}{2*-16}=\frac{-150+26\sqrt{33}}{-32}
$