Answer:
[tex]S=\{ \frac{-6}{5}+\frac{2}{5}i, \frac{-6}{5}-\frac{2}{5}i\}[/tex]
Step-by-step explanation:
[tex]5x^2+12x+8 =0[/tex]
We are going to use the general formula to find the solution. That formula is
[tex]x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
where a= 5, b=12 and c=8. Then
[tex]x = \frac{-12\pm\sqrt{144-4(5)(8)}}{10}[/tex]
[tex]x = \frac{-12\pm\sqrt{144-4(5)(8)}}{10}[/tex]
[tex]x = \frac{-12\pm\sqrt{-16}}{10}[/tex]
[tex]x = \frac{-12\pm4i}{10}[/tex]
[tex]x_1 = \frac{-12+4i}{10}[/tex]
[tex]x_1 = \frac{-6}{5}+\frac{2}{5}i[/tex]
[tex]x_2 = \frac{-12-4i}{10}[/tex]
[tex]x_2 = \frac{-6}{5}-\frac{2}{5}i[/tex].
Then, the solution set is [tex]S=\{ \frac{-6}{5}+\frac{2}{5}i, \frac{-6}{5}-\frac{2}{5}i\}[/tex]
