For this case we have the following polynomial:
[tex]3x ^ 2 - 4x + 5 = 0
[/tex]
Using the resolver we have:
[tex]x = \frac{-b +/- \sqrt{b^2 - 4ac}}{2a} [/tex]
Substituting values we have:
[tex]x = \frac{-(-4) +/- \sqrt{(-4)^2 - 4*3*5}}{2*3}
[/tex]
[tex]x = \frac{4 +/- \sqrt{16 - 60}}{6} [/tex]
[tex]x = \frac{4 +/- \sqrt{-44}}{6} [/tex]
[tex]x = \frac{4 +/- \sqrt{-4*11}}{6} [/tex]
[tex]x = \frac{4 +/- 2\sqrt{-11}}{6} [/tex]
[tex]x = \frac{2 +/- \sqrt{-11}}{3} [/tex]
[tex]x = \frac{2 +/- i\sqrt{11}}{3}[/tex]
Answer:
We observe that the solution to the quadratic equation are two imaginary roots.
