0.24x^2+104x=0

Solution for 0.24x^2+104x=0 equation:

0.24x^2+104x=0

a = 0.24; b = 104; c = 0;
Δ = b2-4ac
Δ = 1042-4·0.24·0
Δ = 10816
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10816}=104$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(104)-104}{2*0.24}=\frac{-208}{0.48}
=-433+0.16/0.48
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(104)+104}{2*0.24}=\frac{0}{0.48}
=0
$