4.905x^2+12x=0

Solution for 4.905x^2+12x=0 equation:

4.905x^2+12x=0

a = 4.905; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·4.905·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*4.905}=\frac{-24}{9.81}
=-2+3.1739130434783/7.1086956521739
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*4.905}=\frac{0}{9.81}
=0
$