(3x)^2+2x+0=0

Solution for (3x)^2+2x+0=0 equation:

(3x)^2+2x+0=0

We add all the numbers together, and all the variables

3x^2+2x=0

a = 3; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·3·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*3}=\frac{-4}{6}
=-2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*3}=\frac{0}{6}
=0
$