Answer:
[tex]\large\boxed{-\dfrac{1}{3}-\dfrac{\sqrt{14}}{3}i,\ -\dfrac{1}{3}+\dfrac{\sqrt{14}}{3}i}[/tex]
Step-by-step explanation:
Use
[tex]ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]3x^2+2x+5=0\\\\a=3,\ b=2,\ c=5\\\\b^2-4ac=2^2-4(3)(5)=4-60=-56\\\\\sqrt{b^2-4ac}=\sqrt{-56}=\sqrt{(4)(-14)}=\sqrt4\cdot\sqrt{-14}=2\sqrt{-14}[/tex]
Use
[tex]i=\sqrt{-1}[/tex]
[tex]\sqrt{-14}=\sqrt{(-1)(14)}=\sqrt{-1}\cdot\sqrt{14}=i\sqrt{14}[/tex]
Therefore:
[tex]x=\dfrac{-2\pm 2i\sqrt{14}}{2(3)}=-\dfrac{2}{6}\pm\dfrac{2i\sqrt{14}}{6}=-\dfrac{1}{3}\pm \dfrac{\sqrt{14}}{3}i[/tex]
