3x^2+6(2x)=0

Solution for 3x^2+6(2x)=0 equation:

3x^2+6(2x)=0

a = 3; b = 62; c = 0;
Δ = b2-4ac
Δ = 622-4·3·0
Δ = 3844
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3844}=62$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(62)-62}{2*3}=\frac{-124}{6}
=-20+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(62)+62}{2*3}=\frac{0}{6}
=0
$