4.9t(^2)+2t-20=0

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

4.9t(^2)+2t-20=0

a = 4.9; b = 2; c = -20;
Δ = b2-4ac
Δ = 22-4·4.9·(-20)
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6\sqrt{11}}{2*4.9}=\frac{-2-6\sqrt{11}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6\sqrt{11}}{2*4.9}=\frac{-2+6\sqrt{11}}{9.8}
$