4.9t^2-1603t+8=0

Solution for 4.9t^2-1603t+8=0 equation:

4.9t^2-1603t+8=0

a = 4.9; b = -1603; c = +8;
Δ = b2-4ac
Δ = -16032-4·4.9·8
Δ = 2569452.2
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{-(-1603)-\sqrt{2569452.2}}{2*4.9}=\frac{1603-\sqrt{2569452.2}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1603)+\sqrt{2569452.2}}{2*4.9}=\frac{1603+\sqrt{2569452.2}}{9.8}
$