Radical Equation transformed by simple substitution to first degree Linear equation

Exercise:

Solve in $\mathbb{R},$ $\sqrt{x-2+\sqrt{2x-5}}+\sqrt{x+2+3\sqrt{2x-5}}=7\sqrt{2}$

Solution: Let $u=x-2\Leftrightarrow x=u+2$

So $\sqrt{u+\sqrt{2u-1}}+\sqrt{u+4+3\sqrt{2u-1}}=7\sqrt{2}$

Put ${{w}^{2}}=2u-1\Leftrightarrow u=\frac{{{w}^{2}}+1}{2}$

$\sqrt{\frac{{{w}^{2}}+1}{2}+w}+\sqrt{\frac{{{w}^{2}}+1}{2}+4+3w}=7\sqrt{2}$

$\Rightarrow \frac{\sqrt{{{w}^{2}}+1+2w}+\sqrt{{{w}^{2}}+1+8+6w}}{\sqrt{2}}=7\sqrt{2}$

$\Rightarrow \sqrt{{{\left( w+1 \right)}^{2}}}+\sqrt{{{\left( w+3 \right)}^{2}}}=14$

$\Rightarrow w+1+w+3=14\Leftrightarrow 2w+4=14\Leftrightarrow 2w=10\Leftrightarrow w=5$

By back word substitution to get $2u={{w}^{2}}+1=25+1=26\Leftrightarrow u=13$

Hence $x=13+2=15$