4.9t^2+4t-80=0

Solution for 4.9t^2+4t-80=0 equation:

4.9t^2+4t-80=0

a = 4.9; b = 4; c = -80;
Δ = b2-4ac
Δ = 42-4·4.9·(-80)
Δ = 1584
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:

$\sqrt{\Delta}=\sqrt{1584}=\sqrt{144*11}=\sqrt{144}*\sqrt{11}=12\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12\sqrt{11}}{2*4.9}=\frac{-4-12\sqrt{11}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12\sqrt{11}}{2*4.9}=\frac{-4+12\sqrt{11}}{9.8}
$