# How do you complete the square to find the vertex for y=x^2-3x+2?

To complete the square half the coefficient of x, that is $- \frac{3}{2}$, then square it, that is $\frac{9}{4}$. Now add and subtract it to the expression as follows:
y=${x}^{2} - 3 x + \frac{9}{4} - \frac{9}{4} + 2$
=${\left(x - \frac{3}{2}\right)}^{2} - \frac{1}{4}$. That is how 'complete the square' is done.
Vertex is $\left(\frac{3}{2} , - \frac{1}{4}\right)$