The given equation is $(a-b) x^{2}+5(a+b) x-2(a-b)=0$.

$\begin{array}{l}
\therefore D=[5(a+b)]^{2}-4 \times(a-b) \times[-2(a-b)] \\
=25(a+b)^{2}+8(a-b)^{2}
\end{array}$

Since a and $\mathrm{b}$ are real and $a \neq b$, so $(a-b)^{2}>0$ and
$(a+b)^{2}>0$.

$\therefore 8(a-b)^{2}>0$

(1) (Product of two positive numbers is always positive)

Also, $25(a+b)^{2}>0$

(2) (Product of two positive numbers is always positive)

Adding (1) and (2), we get

$25(a+b)^{2}+8(a-b)^{2}>0$ (Sum of two positive numbers is always positive)

$\Rightarrow D>0$

Hence, the roots of the given equation are real and unequal.