The vertex and the focus of the parabola $2x^2+5y-6x+1=0$ respectively, are

Given:

$2x^2+5y-6x+1=0$

$\Rightarrow x^2-3x+\frac{1}{2}=-\frac{5}{2}y$

$\Rightarrow x^2-3x+\frac{9}{4}+\frac{1}{2}-\frac{9}{4}=-\frac{5}{2}y$

$\Rightarrow {\left(x-\frac{3}{2}\right)}^2=-\left(\frac{10}{4}\right)y+\frac{7}{4}$

$\Rightarrow {\left(x-\frac{3}{2}\right)}^2=-\frac{10}{4}\left(y-\frac{7}{10}\right)$

$\Rightarrow {\left(x-\frac{3}{2}\right)}^2=-\frac{5}{2}\left(y-\frac{7}{10}\right)$

$\Rightarrow {\left(x-\frac{3}{2}\right)}^2=-4\times \frac{5}{8}\times \left(y-\frac{7}{10}\right)$

Vertex is $\left(\frac{3}{2},\frac{7}{10}\right)$.

Focus of the parabola $X^2=-4AY$ is $\left(X=0,Y=-A\right)$

So, $y-\frac{7}{10}=-\frac{5}{8}\Rightarrow y=\frac{3}{40}$

Hence, focus of given parabola is

$\left(\frac{3}{2},\frac{3}{40}\right)$.