-4.9t^2+400t-80=0

Solution for -4.9t^2+400t-80=0 equation:

-4.9t^2+400t-80=0

a = -4.9; b = 400; c = -80;
Δ = b2-4ac
Δ = 4002-4·(-4.9)·(-80)
Δ = 158432
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{158432}=\sqrt{16*9902}=\sqrt{16}*\sqrt{9902}=4\sqrt{9902}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(400)-4\sqrt{9902}}{2*-4.9}=\frac{-400-4\sqrt{9902}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(400)+4\sqrt{9902}}{2*-4.9}=\frac{-400+4\sqrt{9902}}{-9.8}
$