If $3{\mathrm{p}}^2=5\mathrm{p}+2$ and $3{\mathrm{q}}^2=5\mathrm{q}+2$ where $\mathrm{p}\ne \mathrm{q}$, then the equation whose roots are $3\mathrm{p}-2\mathrm{q}$ and $3\mathrm{q}-2\mathrm{p}$ is

Given roots are $3\mathrm{p}-2\mathrm{q} \mathrm{and} 3\mathrm{q}-2\mathrm{p}.$

Sum of roots $=(3\mathrm{p}-2\mathrm{q})+(3\mathrm{q}-2\mathrm{p})=(\mathrm{p}+\mathrm{q})=\frac{5}{3}$

Product of roots $=(3\mathrm{p}-2\mathrm{q})(3\mathrm{q}-2\mathrm{p})$

$=9\mathrm{pq}-6{\mathrm{q}}^2-6{\mathrm{p}}^2+4\mathrm{pq}=13\mathrm{pq}-2(3{\mathrm{p}}^2+3{\mathrm{q}}^2)$

$=13\left(\frac{-2}{3}\right)-2(5\mathrm{p}+2+5\mathrm{q}+2)$

$=13\left(\frac{-2}{3}\right)-2\left[5\right(\frac{5}{3})+4]$

$=\frac{-26}{3}-2[\frac{25}{3}+4]=\frac{-100}{3}$

Hence, equation is $3{\mathrm{x}}^2-5\mathrm{x}-100=0$