If x = 5 + 2 , then the value of 2x^2 - 3x - 23x^2 - 4x - 3 is equal to?

Answer: Option C

$$x=\sqrt{5}+2$$
Rationalization of denominator.
$$\begin{array}{rl}& \Rightarrow \frac{1}{x}=\frac{1}{\sqrt{5}+2}\times \frac{\sqrt{5}-2}{\sqrt{5}-2}\\ & \Rightarrow \frac{1}{x}=\frac{\sqrt{5}-2}{5-4}\\ & \Rightarrow \frac{1}{x}=\sqrt{5}-2\\ & \Rightarrow x-\frac{1}{x}=\sqrt{5}+2-\sqrt{5}+2\\ & \Rightarrow x-\frac{1}{x}=4\\ & \therefore \frac{2x^2-3x-2}{3x^2-4x-3}\end{array}$$
Multiply by $$\frac{1}{x}$$ in numerator and denominator
$$\begin{array}{rl}& =\frac{\frac{2x^2}{x}-\frac{3x}{x}-\frac{2}{x}}{\frac{3x^2}{x}-\frac{4x}{x}-\frac{3}{x}}\\ & =\frac{2x-\frac{2}{x}-3}{3x-\frac{3}{x}-4}\\ & =\frac{2\left(x-\frac{1}{x}\right)-3}{3\left(x-\frac{1}{x}\right)-4}\\ & =\frac{2\times 4-3}{3\times 4-4}\\ & =\frac{8-3}{12-4}\\ & =\frac{5}{8}\\ & =0.625\end{array}$$