20+15t-4.905t^2=0

Solution for 20+15t-4.905t^2=0 equation:

20+15t-4.905t^2=0

a = -4.905; b = 15; c = +20;
Δ = b2-4ac
Δ = 152-4·(-4.905)·20
Δ = 617.4
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{-(15)-\sqrt{617.4}}{2*-4.905}=\frac{-15-\sqrt{617.4}}{-9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{617.4}}{2*-4.905}=\frac{-15+\sqrt{617.4}}{-9.81}
$