Solve the following equation and verify your answer:

$\frac{(x+2)(2 x-3)-2 x^{2}+6}{x-5}=2$

or $\frac{2 x^{2}+x-6-2 x^{2}+6}{x-5}=2$

or $\frac{\mathrm{x}}{\mathrm{x}-5}=2$

or $2 \mathrm{x}-10=\mathrm{x}[$ After c ross multiplication $]$

or $2 \mathrm{x}-\mathrm{x}=10$

or $\mathrm{x}=10$

Thus, $x=10$ is the solution of the given equation.

Check :

Substituting $x=10$ in the given equation, we get:

L. H.S. $=\frac{(10+2)(2 \times 10-3)-2 \times 10^{2}+6}{10-5}=\frac{12 \times 17-200+6}{5}=\frac{10}{5}=2$

R.H.S. $=2$

$\therefore$ L.H.S. $=$ R.H.S. for $\mathrm{x}=10$.