Finding $y$ from $y'=\sqrt{5x+2y-3}$. I'd love your help with finding the function $y$ from the following differential equation:
$y'=\sqrt{5x+2y-3}$.
I tried to use $z=5x+2-3$, so $z'=5+2y'$ , and $y'=\frac{z'}{2}-2.5$
and from the equation $\frac{z'}{2}-2.5=y'=\sqrt z$, and then $z'=2 \sqrt z+5$, so $\frac{dz}{2 \sqrt z +5} = dx$, but using integration here is difficult and  won't lead me to $y$.
I tries to use substitution in other ways like $z=\sqrt{5x+2y-3}$, or $z^2= 5x+2y-3$ but then again I got stuck in the middle..
Any suggestion?
Thanks a lot

Integration here is simple
$$
\int\frac{dz}{2\sqrt{z}+5}=\{t=\sqrt{z}\}=\int\frac{2t dt}{2t+5}=\int\left(1-\frac{5}{2t+5}\right)dt=t-\frac{5}{2}\int\frac{d(2t+5)}{2t+5}=
$$
$$
t-\frac{5}{2}\ln(2t+5)+C=\sqrt{z}-\frac{5}{2}\ln(2\sqrt{z}+5)+C
$$
