Answer: [tex]\bold{-\dfrac{1}{2}\pm \dfrac{\sqrt5}{10}}i[/tex]
Step-by-step explanation:
[tex]10y^2+10y+3=0\qquad \rightarrow \quad a=10,\ b=10,\ c=3\\\\y=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\y=\dfrac{-10\pm \sqrt{10^2-4(10)(3)}}{2(10)}\\\\\\.\ =\dfrac{-10\pm \sqrt{100-120}}{2(10)}\\\\\\.\ =\dfrac{-10\pm \sqrt{-20}}{2(10)}\\\\\\.\ =\dfrac{-10\pm 2i\sqrt5}{2(10)}\\\\\\.\ =\dfrac{-5\pm i\sqrt5}{10}\\\\\\.\ =-\dfrac{1}{2}\pm \dfrac{\sqrt5}{10}i[/tex]
