$$\max_{x_1,x_2} f(x_1,x_2) \text{ subject to }g(x_1,x_2)=0.$$ So, for critical points, form Lagrangian$$\mathcal L(x_1,x_2,\lambda)\equiv f(x_1,x_2)-\lambda g(x_1,x_2).$$ and then, $$\dfrac{\partial \mathcal L}{\partial x_1}=\dfrac{\partial f(x_1,x_2)}{\partial x_1}-\lambda\dfrac{\partial g(x_1,x_2)}{\partial x_1}=0$$ $$\dfrac{\partial \mathcal L}{\partial x_2}=\dfrac{\partial f(x_1,x_2)}{\partial x_2}-\lambda\dfrac{\partial g(x_1,x_2)}{\partial x_2}=0$$$$\dfrac{\partial \mathcal L}{\partial \lambda}=g(x_1,x_2)=0.$$ Now let $g(x_1,x_2)=p(x_1,x_2)-b=0$, for some function $p$ and constant $b$. But if we interpret $g$ as $b-p(x_1,x_2)=0$, critical points for $x_1$ and $x_2$ remain the same but $\lambda$ reverses the sign.
So, if the constraint is given as $p(x_1,x_2)=b$ is given, which $\lambda$ is "correct"?(i.e., how should we write constraint?)(Sign of Lagrangian multiplier matters very much in economics.)