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!

Denote $t:=\sqrt{45x^2-30x+1}$. Then we observe that
$$t^2=45x^2-30x+1=-5(7+6x-9x^2)+36.$$
As a result, it follows that $$t=-\frac{t^2-36}{5}\implies t^2+5t-36=0\implies(t+9)(t-4)=0.$$
Since $t\geq 0$, it follows that $t=4$. Therefore, $$45x^2-30x+1=16\implies45x^2-30x-15=0\implies 3x^2-2x-1=0.$$
Hence $(3x+1)(x-1)=0$. Either $x=1$ or $x=-1/3$.
