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

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

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

We get rid of parentheses

2x^2+6x+3=0

a = 2; b = 6; c = +3;
Δ = b2-4ac
Δ = 62-4·2·3
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{3}}{2*2}=\frac{-6-2\sqrt{3}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{3}}{2*2}=\frac{-6+2\sqrt{3}}{4}
$