42x^2+20x+4x=0

Solution for 42x^2+20x+4x=0 equation:

42x^2+20x+4x=0

We add all the numbers together, and all the variables

42x^2+24x=0

a = 42; b = 24; c = 0;
Δ = b2-4ac
Δ = 242-4·42·0
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24}{2*42}=\frac{-48}{84}
=-4/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24}{2*42}=\frac{0}{84}
=0
$