Just wanted to check if this was right before I proceed

f(x,y)=$2x^3 + 6xy^2 - 3y^3 - 150x$

which gives 

$\frac{∂f}{∂x}$ = $6x^2 + 6y^2 -150$

Then doing the same with y gives 

$\frac{∂f}{∂y}$ = $12xy -9y^2$

To find the stationary points, I have to make the derivatives 0 which gives me

$\frac{∂f}{∂x}$ = $6x^2 + 6y^2 -150$ = $0$
and $\frac{∂f}{∂y}$ = $12xy -9y^2$ = $0$

rearranging $\frac{∂f}{∂x}$= $0$ gives me $y^2 = -x^2 + 25$

I proceed to sub this into $\frac{∂f}{∂y}$ = $12xy -9y^2$ = $0$

This gives me $-12x^2 + 60x + 9x^2 - 225 = 0$

and putting it into quadratic eqtn and then factorising, I get $x=5$ and $x=15$

This is where I get confused, do these numbers sound right and if so, do I place the x coordinates into the original eqtn to get y coordinates? 

how many stationary points in total?