Standard form
Coeficients are: a = 3; b = 4; c = 2;
Because \(b^{2} -4ac = -8\) is negative we get two Complex solutions;
We can use the special Quadratic Formula
x = (-b + or - i\(\sqrt{ b^{2}-4ac }\))/(2a); where i = \(\sqrt{-1} ;\)
Then, x1 = (-4 + i2\(\sqrt{2} )/6 = 2(-2 + i \sqrt{2} )/6 = (-2+i \sqrt{2} )/3;\);
In similar way, we obtain x2 = (-2-i \sqrt{2} )/3;[/tex];
