-4.905t^2+50t-90=0

Solution for -4.905t^2+50t-90=0 equation:

-4.905t^2+50t-90=0

a = -4.905; b = 50; c = -90;
Δ = b2-4ac
Δ = 502-4·(-4.905)·(-90)
Δ = 734.2
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-\sqrt{734.2}}{2*-4.905}=\frac{-50-\sqrt{734.2}}{-9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+\sqrt{734.2}}{2*-4.905}=\frac{-50+\sqrt{734.2}}{-9.81}
$