4.9t^2-20t+20=0

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

4.9t^2-20t+20=0

a = 4.9; b = -20; c = +20;
Δ = b2-4ac
Δ = -202-4·4.9·20
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{2}}{2*4.9}=\frac{20-2\sqrt{2}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{2}}{2*4.9}=\frac{20+2\sqrt{2}}{9.8}
$