-4.9t^2+39.2t=0

Solution for -4.9t^2+39.2t=0 equation:

-4.9t^2+39.2t=0

a = -4.9; b = 39.2; c = 0;
Δ = b2-4ac
Δ = 39.22-4·(-4.9)·0
Δ = 1536.64
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{-(39.2)-\sqrt{1536.64}}{2*-4.9}=\frac{-39.2-\sqrt{1536.64}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39.2)+\sqrt{1536.64}}{2*-4.9}=\frac{-39.2+\sqrt{1536.64}}{-9.8}
$