150t-16.1t^2+50=0

Solution for 150t-16.1t^2+50=0 equation:

150t-16.1t^2+50=0

a = -16.1; b = 150; c = +50;
Δ = b2-4ac
Δ = 1502-4·(-16.1)·50
Δ = 25720
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:

$\sqrt{\Delta}=\sqrt{25720}=\sqrt{4*6430}=\sqrt{4}*\sqrt{6430}=2\sqrt{6430}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(150)-2\sqrt{6430}}{2*-16.1}=\frac{-150-2\sqrt{6430}}{-32.2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(150)+2\sqrt{6430}}{2*-16.1}=\frac{-150+2\sqrt{6430}}{-32.2}
$