We have, $36{x^2} + 60x + 25 = \frac{{35}}{{3{x^2} + 5x + 2}}$

$ \Rightarrow 12(3{x^2} + 5x) + 25 = \frac{{35}}{{(3{x^2} + 5x) + 2}}$

Let, $3{x^2} + 5x = t$

So, $12t + 25 = \frac{{35}}{{t + 2}}$

$ \Rightarrow 12{t^2} + 49t + 15 = 0$

$ \Rightarrow t = \frac{{ – 49 \pm \sqrt {{{49}^2} – 720} }}{{24}} = \frac{{ – 49 \pm 41}}{{24}} = – \frac{1}{3}, – \frac{{15}}{4}$

Now, $3{x^2} + 5x = – \frac{1}{3}, – \frac{{15}}{4}$

$9{x^2} + 15x + 1 = 0$ equation yields, $x = \frac{{ – 5 \pm \sqrt {21} }}{6}$

The other equation $12{x^2} + 20x + 15 = 0$ has discriminant < 0, hence no real solutions.