Question

$$(\frac{2x+7}{4})^{2}=\frac{48u^{2}+529}{16}$$

Solve for u (complex solution)

$u=-\frac{i\sqrt{24-3x}\sqrt{x+15}}{6}$
$u=\frac{i\sqrt{24-3x}\sqrt{x+15}}{6}$

Solve for u

$u=\frac{\sqrt{3\left(x-8\right)\left(x+15\right)}}{6}$
$u=-\frac{\sqrt{3\left(x-8\right)\left(x+15\right)}}{6}\text{, }x\leq -15\text{ or }x\geq 8$

Solve for x

$x=\frac{\sqrt{48u^{2}+529}-7}{2}$
$x=\frac{-\sqrt{48u^{2}+529}-7}{2}$