The quadratic formula used for quadratic equation \(\displaystyle{a}{x}^{{2}}+{b}{x}+{c}={0}\) is,

\(\displaystyle{x}=-\frac{{{b}\pm\sqrt{{{b}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}\)

Apply the quadratic formula in \(\displaystyle{6}{x}^{{2}}-{x}-{15}={0}\) then

\(\displaystyle{x}=\frac{{-{1}{\left(-{1}\right)}\pm\sqrt{{{\left(-{1}\right)}^{{2}}-{4}\times{6}\times{\left(-{15}\right)}}}}}{{{2}\times{6}}}\)

\(\displaystyle{x}=\frac{{{1}\pm\sqrt{{{1}+{160}}}}}{{12}}\)

\(\displaystyle{x}=\frac{{{1}\pm\sqrt{{{361}}}}}{{6}}\)

\(\displaystyle{x}=\frac{{{1}+{19}}}{{6}}\)

\(\displaystyle{x}=\frac{{5}}{{3}},\frac{{3}}{{2}}\)

\(\displaystyle{x}=-\frac{{{b}\pm\sqrt{{{b}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}\)

Apply the quadratic formula in \(\displaystyle{6}{x}^{{2}}-{x}-{15}={0}\) then

\(\displaystyle{x}=\frac{{-{1}{\left(-{1}\right)}\pm\sqrt{{{\left(-{1}\right)}^{{2}}-{4}\times{6}\times{\left(-{15}\right)}}}}}{{{2}\times{6}}}\)

\(\displaystyle{x}=\frac{{{1}\pm\sqrt{{{1}+{160}}}}}{{12}}\)

\(\displaystyle{x}=\frac{{{1}\pm\sqrt{{{361}}}}}{{6}}\)

\(\displaystyle{x}=\frac{{{1}+{19}}}{{6}}\)

\(\displaystyle{x}=\frac{{5}}{{3}},\frac{{3}}{{2}}\)