Respuesta :
The quadratic formula states that the solutions x1 and x2 of a quadratic function in the form y = ax^2 + bx + c is equal to:
[tex]\begin{gathered} x_1=\frac{-b+\sqrt[]{b^2-4ac}}{2a} \\ x_2=\frac{-b-\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]So, using this formula with the values a = 3, b = 4 and c = 2, we have that:
[tex]\begin{gathered} x_1=\frac{-4+\sqrt[]{4^2-4\cdot3\cdot2}}{2\cdot3}=\frac{-4+\sqrt[]{16-24}}{6}=\frac{-4+\sqrt[]{-8}}{6} \\ x_1=\frac{-4+\sqrt[]{2^2\cdot(-2)}}{6}=\frac{-4+2\cdot\sqrt[]{-2}}{6}=\frac{-2+\sqrt[]{-2}}{3}=-\frac{2}{3}+i\cdot\frac{\sqrt[]{2}}{3} \\ x_2=\frac{-4-\sqrt[]{-8}}{6}=\frac{-2-\sqrt[]{-2}}{3}=-\frac{2}{3}-i\cdot\frac{\sqrt[]{2}}{3} \end{gathered}[/tex]Since the zeros have a complex part, the solutions are imaginary.
