Solve the equation $\sqrt{45x^2-30x+1}=7+6x-9x^2$ Solve the equation $$\sqrt{45x^2-30x+1}=7+6x-9x^2.$$
So we have $\sqrt{45x^2-30x+1}=7+6x-9x^2\iff \begin{cases}7+6x-9x^2\ge0\\45x^2-30x+1=(7+6x-9x^2)^2\end{cases}.$ The inequality gives $x\in\left[\dfrac{1-2\sqrt{2}}{3};\dfrac{1+2\sqrt{2}}{3}\right].$ I am not sure how to deal with the equation. Thank you in advance!

Let $u=9x^2-6x$ and rewrite the equation as
$$\sqrt{5u+1}=7-u$$
Squaring both sides gives $5u+1=49-14u+u^2$, or
$$u^2-19u+48=(u-16)(u-3)=0$$
We see that $u=16$ is not a solution, since $\sqrt{81}\not=-9$. This leaves us with $u=3$, which is a valid solution, since $\sqrt{16}=4$, and from this we have
$$9x^2-6x=3\implies3x^2-2x-1=(x-1)(3x+1)=0$$
which gives $x=1$ and $x=-1/3$ as the complete solution set.
Remark: What makes this work so nicely is that $45:30=9:6$. If the coefficients of $x^2$ and $x$ on the two sides hadn't been in proportion, the solution would have been much more involved.
