-4.905t^2+11t+90=0

Solution for -4.905t^2+11t+90=0 equation:

-4.905t^2+11t+90=0

a = -4.905; b = 11; c = +90;
Δ = b2-4ac
Δ = 112-4·(-4.905)·90
Δ = 1886.8
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{-(11)-\sqrt{1886.8}}{2*-4.905}=\frac{-11-\sqrt{1886.8}}{-9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{1886.8}}{2*-4.905}=\frac{-11+\sqrt{1886.8}}{-9.81}
$