1. Move everything to the right side

$(x+c)(x+d)-(2 x+c+d)=0$,

$x^{2}+c x+d x+c d-2 x-c-d=0$,

$x^{2}+(c+d-2) x+c d-c-d=0$.

Find the discriminant of the quadratic equation

$D=(c+d-2)^{2}-4(cd-c-d)=c^{2}+d^{2}+4+2cd-4c-4d-4cd+4c+4d=$ $=c^{2}+d^{2}-2cd+4=(c-d)^{2}+4>0$ Therefore, the equation has two distinct roots.
Solution 2. Consider $\mathrm{f}(\mathrm{x})=(\mathrm{x}+\mathrm{c})(\mathrm{x}+\mathrm{d})-(2 \mathrm{x}+\mathrm{c}+\mathrm{d})$, then the equation from the condition will take the form $\mathrm{f}(\mathrm{x})=0$. Notice that $\mathrm{f}(-\mathrm{c})=\mathrm{c}-\mathrm{d}$ and $\mathrm{f}(-\mathrm{d})=\mathrm{d}-\mathrm{c}$. Since $\mathrm{c} \neq \mathrm{d}$, in one of these points the value of this reduced quadratic polynomial is negative. Therefore, the quadratic polynomial has exactly two roots.

Remark. In the first solution, the condition $\mathrm{c} \neq \mathrm{d}$ is not used.