-4.9t^2+19.6t+90=0

Solution for -4.9t^2+19.6t+90=0 equation:

-4.9t^2+19.6t+90=0

a = -4.9; b = 19.6; c = +90;
Δ = b2-4ac
Δ = 19.62-4·(-4.9)·90
Δ = 2148.16
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{-(19.6)-\sqrt{2148.16}}{2*-4.9}=\frac{-19.6-\sqrt{2148.16}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19.6)+\sqrt{2148.16}}{2*-4.9}=\frac{-19.6+\sqrt{2148.16}}{-9.8}
$