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(\dfrac{6}{2}\right)^2 $ after the $ 6x $ and $ \left(\dfrac{10}{2}\right)^2 $ after the $ 10y $. So we get $$9\left(x^2-6x+9\right)+4\left(y^2+10y+25\right)=-145+9\cdot9+4\cdot25=-145+81+100=36$$Dividing both sides by $ 36 $ gives $$\dfrac{\left(x-3\right)^2}{2^2}+\dfrac{\left(y+5\right)^2}{3^2}=1.$$Hence, $ h + k+ a + b = 3 +(-5) + 2 + 3 = \boxed{3} $.