jmc

First, group the terms as follows: $$\left(9x^2-54x\right)+\left(4y^2+40y\right)=-145$$Factoring out the coefficients of $x^2$ and $y^2$ gives $$9\left(x^2-6x\right)+4\left(y^2+10y\right)=-145$$To complete the square, we need to add ${\left(\frac{6}{2}\right)}^2$ after the $6x$ and ${\left(\frac{10}{2}\right)}^2$ after the $10y$. So we get $$9\left(x^2-6x+9\right)+4\left(y^2+10y+25\right)=-145+9\cdot 9+4\cdot 25=-145+81+100=36$$Dividing both sides by $36$ gives $$\frac{{\left(x-3\right)}^2}{2^2}+\frac{{\left(y+5\right)}^2}{3^2}=1.$$Hence, $h+k+a+b=3+(-5)+2+3=\boxed{3}.$