4.9t^2-20t-80=0

Solution for 4.9t^2-20t-80=0 equation:

4.9t^2-20t-80=0

a = 4.9; b = -20; c = -80;
Δ = b2-4ac
Δ = -202-4·4.9·(-80)
Δ = 1968
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:

$\sqrt{\Delta}=\sqrt{1968}=\sqrt{16*123}=\sqrt{16}*\sqrt{123}=4\sqrt{123}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{123}}{2*4.9}=\frac{20-4\sqrt{123}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{123}}{2*4.9}=\frac{20+4\sqrt{123}}{9.8}
$