Solve a^2x^2-3abx+2b^2 by completing square

The given equation is:
a²x² - 3abx + 2b² = 0
{tex}\Rightarrow{/tex} a²x² - 3abx = -2b²
Adding {tex}\left( \frac { 3 b } { 2 } \right) ^ { 2 }{/tex} on both sides, we have,
{tex}(ax)^2-2\times(ax)\times\frac{3b}2+\left(\frac{3b}2\right)^2=-2b^2+\left(\frac{3b}2\right)^2{/tex}
{tex}(ax-\frac{3b}2)^2=-2b^2+(\frac{3b}2)^2{/tex}
{tex}(ax-\frac{3b}2)^2=\frac{b^2}4{/tex} 
{tex}\Rightarrow \quad \left( a x - \frac { 3 b } { 2 } \right) = \pm \frac { b } { 2 }{/tex} (By taking square root on both sides)
So either, {tex} \left( a x - \frac { 3 b } { 2 } \right) = \frac { b } { 2 }{/tex} or {tex}\left( a x - \frac { 3 b } { 2 } \right) = \frac { - b } { 2 }{/tex}
{tex}\Rightarrow \quad a x = \left( \frac { b } { 2 } + \frac { 3 b } { 2 } \right) = \frac { 4 b } { 2 }{/tex}= 2b,
thus, x = 2b/a
or ax = {tex}\left( \frac { - b } { 2 } + \frac { 3 b } { 2 } \right) = \frac { 2 b } { 2 }{/tex} = b
thus, x = b/a
So, x = {tex}\frac { 2 b } { a }{/tex} or x = {tex}\frac { b } { a }{/tex}.
Hence, the roots of the given equation are {tex}\frac { 2 b } { a }{/tex} and {tex}\frac { b } { a }{/tex}.