4.9t^2+50t+2=0

Solution for 4.9t^2+50t+2=0 equation:

4.9t^2+50t+2=0

a = 4.9; b = 50; c = +2;
Δ = b2-4ac
Δ = 502-4·4.9·2
Δ = 2460.8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-\sqrt{2460.8}}{2*4.9}=\frac{-50-\sqrt{2460.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+\sqrt{2460.8}}{2*4.9}=\frac{-50+\sqrt{2460.8}}{9.8}
$