You can diagonalise the equation $2a(a+1)=3b(b+1)$ by wiring it as $$2(2a+1)^2+1=3(2b+1)^2.$$

To parametrise all rational solutions of $2a(a+1)=3b(b+1)$ you can now homogenise $$2(2a+1)^2+1=3(2b+1)^2$$ into $$2(2a+1)^2+c^2=3(2b+1)^2$$ and define $$x:=2a+1,y:=c^2,z:=2b+1,$$ so that $$2(2a+1)^2+c^2=3(2b+1)^2$$ becomes $$2x^2+y^2-3z^2=0.$$ This is a diagonal conic with the rational point $(x_0,y_0,z_0)=(1,1,1)$ and you can find the parametrisation of all solutions before Theorem 2.3 these notes.

If you are only interested in the integer solutions of $2a(a+1)=3b(b+1)$ the you could try to do some arithmetic in the number field $\mathbb{Q}(\sqrt{-6})$ by noting that $$N_{\mathbb{Q}(\sqrt{-6})/\mathbb{Q}}(2(2a+1)+\sqrt{-6}(2b+1))=2((2a+1)^2-3(2b+1)^2)=2.$$

You can diagonalise the equation $2a(a+1)=3b(b+1)$ by writing it as $$2(2a+1)^2+1=3(2b+1)^2.$$

To parametrise all rational solutions of $2a(a+1)=3b(b+1)$ you can now homogenise $$2(2a+1)^2+1=3(2b+1)^2$$ into $$2(2a+1)^2+c^2=3(2b+1)^2$$ and define $$x:=2a+1,y:=c,z:=2b+1,$$ so that $$2(2a+1)^2+c^2=3(2b+1)^2$$ becomes $$2x^2+y^2-3z^2=0.$$ This is a diagonal conic with the rational point $(x_0,y_0,z_0)=(1,1,1)$ and you can find the parametrisation of all solutions before Theorem 2.3 these notes.

If you are only interested in the integer solutions of $2a(a+1)=3b(b+1)$ the you could try to do some arithmetic in the number field $\mathbb{Q}(\sqrt{-6})$ by noting that $$N_{\mathbb{Q}(\sqrt{-6})/\mathbb{Q}}(2(2a+1)+\sqrt{-6}(2b+1))=2((2a+1)^2-3(2b+1)^2)=2.$$