[tex]\textit{quadratic formula}\\
{{ 3}}x^2{{ +b}}x{{ +10}}=0
\qquad \qquad
x= \cfrac{ - {{ b}} \pm \sqrt { {{ b}}^2 -4{{ a}}{{ c}}}}{2{{ a}}}
\\ \quad \\
meaning\implies x=\cfrac{-b\pm\sqrt{b^2-4(3)(10)}}{2(3)}\qquad now
\\ \quad \\
\cfrac{-b\pm\sqrt{b^2-4(3)(10)}}{2(3)}\to
\begin{cases}
\cfrac{-b+\sqrt{b^2-4(3)(10)}}{2(3)}
\\ \quad \\
\cfrac{-b-\sqrt{b^2-4(3)(10)}}{2(3)}
\end{cases}\impliedby \textit{two roots}[/tex]
[tex]\textit{and their root difference is }4\frac{1}{3}\implies \cfrac{13}{3}\qquad so
\\ \quad \\
\left[ \cfrac{-b+\sqrt{b^2-4(3)(10)}}{2(3)} \right]-\left[ \cfrac{-b-\sqrt{b^2-4(3)(10)}}{2(3)} \right]=\cfrac{13}{3}
\\ \quad \\
\left[ \cfrac{-b+\sqrt{b^2-4(3)(10)}}{6} \right]+\left[ \cfrac{+b+\sqrt{b^2-4(3)(10)}}{6} \right]=\cfrac{13}{3}
\\ \quad \\
\cfrac{-b+\sqrt{b^2-4(3)(10)}+b+\sqrt{b^2-4(3)(10)}}{6}=\cfrac{13}{3}
\\ \quad \\
\cfrac{2\sqrt{b^2-4(3)(10)}}{6}=\cfrac{13}{3}[/tex]
and I'm pretty sure you can take it from there
