-4.9t^2+12t=0

Solution for -4.9t^2+12t=0 equation:

-4.9t^2+12t=0

a = -4.9; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-4.9)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*-4.9}=\frac{-24}{-9.8}
=2+3.1428571428571/7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-4.9}=\frac{0}{-9.8}
=0
$