Elementary Algebra

The only way to check if the equation has two non-real complex solutions or not is to solve the equation: Step 1: Comparing $3x^{2}+2x+1=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find: $a=3$, $b=2$ and $c=1$ Step 2: The quadratic formula is: $x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a, b and c in the formula: $x=\frac{-(2) \pm \sqrt {(2)^{2}-4(3)(1)}}{2(3)}$ Step 4: $x=\frac{-2 \pm \sqrt {4-12}}{6}$ Step 5: $x=\frac{-2 \pm \sqrt {-8}}{6}$ Step 6: $x=\frac{-2 \pm \sqrt {-1\times8}}{6}$ Step 7: $x=\frac{-2 \pm (\sqrt {-1}\times\sqrt {8})}{6}$ Step 8: $x=\frac{-2 \pm (\sqrt {-1}\times\sqrt {4\times2})}{6}$ (Note, $i=\sqrt{-1}$) Step 9: $x=\frac{-2 \pm (i\times 2\sqrt 2)}{6}$ Step 10: $x=\frac{-2 \pm 2i\sqrt 2}{6}$ Step 11: $x=\frac{2(-1 \pm i\sqrt 2)}{6}$ Step 12: $x=\frac{-1 \pm i\sqrt 2}{3}$ Step 13: $x=\frac{-1-i\sqrt 2}{3}$ or $x=\frac{-1 + i\sqrt 2}{3}$ Step 14: Therefore, the solution set is {$\frac{-1 - i\sqrt 2}{3},\frac{-1 + i\sqrt 2}{3}$}. Therefore, the equation does indeed has two non-real complex solutions. This means that the question statement is true.