4.905t^2+8t-20=0

Solution for 4.905t^2+8t-20=0 equation:

4.905t^2+8t-20=0

a = 4.905; b = 8; c = -20;
Δ = b2-4ac
Δ = 82-4·4.905·(-20)
Δ = 456.4
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{-(8)-\sqrt{456.4}}{2*4.905}=\frac{-8-\sqrt{456.4}}{9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+\sqrt{456.4}}{2*4.905}=\frac{-8+\sqrt{456.4}}{9.81}
$