The trigonometric way

Observe that for the validity of the given equation, $|x| < 1$ .

So, Let $x= cos(2 \theta )$ .

The expressions changes to

$5\sqrt{2} (cos \theta + sin \theta) = 6 cos 2\theta + 8 sin 2 \theta$ $\dfrac{1}{\sqrt{2}}(cos \theta + sin \theta) = \dfrac{3}{5} cos 2 \theta + \dfrac{4}{5} sin 2 \theta$

Let, $sin \alpha = \dfrac{3}{5}$ . So, the above expression reduces to

$sin(\dfrac{\pi}{4} + \theta ) = sin( \alpha + 2 \theta)$

$\dfrac{\pi}{4} + \theta = 2 \theta + \alpha$

$\theta = \dfrac{\pi}{4} - \alpha$

Now,

$x= cos (2 \theta) = cos ( \dfrac{\pi}{2} - 2 \alpha ) = sin (2 \alpha)$ $= 2 sin(\alpha) cos (\alpha) =2 \times \dfrac{3}{5} \times \dfrac{4}{5} = \boxed{\dfrac{24}{25}}$ .