Solve each of the following quadratic equations:

We write, $-9(a+b) x=-3(2 a+b) x-3(a+2 b) x$ as $9 x^{2} \times\left(2 a^{2}+5 a b+2 b^{2}\right)=9\left(2 a^{2}+5 a b+2 b^{2}\right) x^{2}=[-3(2 a+b) x] \times[-3(a+2 b) x]$

$\therefore 9 x^{2}-9(a+b) x+\left(2 a^{2}+5 a b+2 b^{2}\right)=0$

$\Rightarrow 9 x^{2}-3(2 a+b) x-3(a+2 b) x+(2 a+b)(a+2 b)=0$

$\Rightarrow 3 x[3 x-(2 a+b)]-(a+2 b)[3 x-(2 a+b)]=0$

$\Rightarrow[3 x-(2 a+b)][3 x-(a+2 b)]=0$

$\Rightarrow 3 x-(2 a+b)=0$ or $3 x-(a+2 b)=0$

$\Rightarrow x=\frac{2 a+b}{3}$ or $x=\frac{a+2 b}{3}$

Hence, $\frac{2 a+b}{3}$ and $\frac{a+2 b}{3}$ are the roots of the given equation.