Answer
$x=\left\{ \dfrac{-3-\sqrt{69}}{6},\dfrac{-3+\sqrt{69}}{6} \right\}$
Work Step by Step
Using the completing the square method, the solutions of the given quadratic equation, $
3x^2+3x=5
,$ is
\begin{array}{l}\require{cancel}
\dfrac{1}{3}\cdot (3x^2+3x)=(5)\cdot\dfrac{1}{3}
\\\\
x^2+x=\dfrac{5}{3}
\\\\
x^2+x+\left( \dfrac{1}{2} \right)^2=\dfrac{5}{3}+\left( \dfrac{1}{2} \right)^2
\\\\
x^2+x+\dfrac{1}{4}=\dfrac{5}{3}+\dfrac{1}{4}
\\\\
\left( x+\dfrac{1}{2}\right)^2=\dfrac{20}{12}+\dfrac{3}{12}
\\\\
\left( x+\dfrac{1}{2}\right)^2=\dfrac{23}{12}
\\\\
x+\dfrac{1}{2}=\pm\sqrt{\dfrac{23}{12}}
\\\\
x+\dfrac{1}{2}=\pm\sqrt{\dfrac{23}{12}\cdot\dfrac{3}{3}}
\\\\
x+\dfrac{1}{2}=\pm\sqrt{\dfrac{69}{36}}
\\\\
x+\dfrac{1}{2}=\pm\dfrac{\sqrt{69}}{6}
\\\\
x=-\dfrac{1}{2}\pm\dfrac{\sqrt{69}}{6}
\\\\
x=-\dfrac{3}{6}\pm\dfrac{\sqrt{69}}{6}
\\\\
x=\dfrac{-3\pm\sqrt{69}}{6}
.\end{array}
Hence, $
x=\left\{ \dfrac{-3-\sqrt{69}}{6},\dfrac{-3+\sqrt{69}}{6} \right\}
.$