Does $-6(x-\frac{1}{2})^2$ = $-\frac{3}{2}(2x-1)^2$?

We have

$$-6\left(x-\frac{1}{2}\right)^2=-\frac 6{\color{blue}4}\cdot \color{blue}4\cdot\left(x-\frac{1}{2}\right)^2=-\frac 3{\color{blue}2}\cdot \color{blue}{2^2}\cdot\left(x-\frac{1}{2}\right)^2=\\=-\frac 32\left(\color{blue}2\cdot x-\color{blue}2\cdot\frac{1}{2}\right)^2=-\frac 32(2x-1)^2$$

Look at my answer, it's quite simple to prove $$-6\left(x-\frac{1}{2}\right)^2=-6\left(\frac{2x-1}{2}\right)^2=-6\frac{\left(2x-1\right)^2}{2^2}=-6\frac{\left(2x-1\right)^2}{4}=-\frac 32(2x-1)^2$$

$$-6\left(x-\frac{1}{2} \right)^2 = -\frac{3}{2}\left(2x-1 \right)^2$$ $$-6\left( x^2 -x +\frac{1}{4}\right) = -\frac{3}{2} \left(4x^2-4x+1 \right)$$ $$-6x^2+6x - \frac{6}{4} = -6x^2+6x -\frac{3}{2} $$