Answer
$x=\left\{ \dfrac{-15-7\sqrt{5}}{10},\dfrac{-15+7\sqrt{5}}{10} \right\}$
Work Step by Step
Using the completing the square method, the solutions of the given quadratic equation, $
5x^2+15x-1=0
,$ is
\begin{array}{l}\require{cancel}
\dfrac{1}{5}\cdot(5x^2+15x-1)=(0)\cdot\dfrac{1}{5}
\\\\
x^2+3x-\dfrac{1}{5}=0
\\\\
x^2+3x=\dfrac{1}{5}
\\\\
x^2+3x+\left( \dfrac{3}{2} \right)^2=\dfrac{1}{5}+\left( \dfrac{3}{2} \right)^2
\\\\
x^2+3x+\dfrac{9}{4}=\dfrac{1}{5}+\dfrac{9}{4}
\\\\
\left(x+\dfrac{3}{2}\right)^2=\dfrac{4}{20}+\dfrac{45}{20}
\\\\
\left(x+\dfrac{3}{2}\right)^2=\dfrac{49}{20}
\\\\
x+\dfrac{3}{2}=\pm\sqrt{\dfrac{49}{20}}
\\\\
x+\dfrac{3}{2}=\pm\sqrt{\dfrac{49}{20}\cdot\dfrac{5}{5}}
\\\\
x+\dfrac{3}{2}=\pm\sqrt{\dfrac{49}{100}\cdot5}
\\\\
x+\dfrac{3}{2}=\pm\dfrac{7}{10}\sqrt{5}
\\\\
x+\dfrac{3}{2}=\pm\dfrac{7\sqrt{5}}{10}
\\\\
x=-\dfrac{3}{2}\pm\dfrac{7\sqrt{5}}{10}
\\\\
x=-\dfrac{15}{10}\pm\dfrac{7\sqrt{5}}{10}
\\\\
x=\dfrac{-15\pm7\sqrt{5}}{10}
.\end{array}
Hence, $
x=\left\{ \dfrac{-15-7\sqrt{5}}{10},\dfrac{-15+7\sqrt{5}}{10} \right\}
.$
