Solve the equation: $\;$ $\sqrt{7^{2x^2 - 5x - 6}} = \left(\sqrt{2}\right)^{3 \log_2 {49}}$
Given equation: $\;\;$ $\sqrt{7^{2x^2 - 5x - 6}} = \left(\sqrt{2}\right)^{3 \log_2 {49}}$
i.e. $\;$ $\left[\left(7\right)^{2x^2 - 5x - 6}\right]^{\frac{1}{2}} = \left[\left(2\right)^{\frac{1}{2}}\right]^{3 \log_2 {7^2}}$
i.e. $\;$ $7^{\frac{2x^2 - 5x - 6}{2}} = 2^{\frac{1}{2} \times 3 \times 2 \log_2 7}$
i.e. $\;$ $7^{\frac{2x^2 - 5x - 6}{2}} = 2^{3 \log_2 7}$
i.e. $\;$ $7^{\frac{2x^2 - 5x - 6}{2}} = 2^{\log_2 {7^3}}$
i.e. $\;$ $7^{\frac{2x^2 - 5x - 6}{2}} = 7^3$
$\implies$ $\dfrac{2x^2 - 5x - 6}{2} = 3$
i.e. $\;$ $2x^2 - 5x - 12 = 0$
i.e. $\;$ $\left(2x + 3\right) \left(x - 4\right) = 0$
$\implies$ $x = \dfrac{-3}{2}$ $\;$ or $\;$ $x = 4$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{-3}{2}, \; 4 \right\}$