hard algebra

As follows:

Solve $\large \sqrt{x+14-8\sqrt{x-2}} + \sqrt{x+23-10\sqrt{x-2}} = 3$

Substitute: $\begin{array}{|rcll|} \hline \mathbf{ y } &=& \mathbf{\sqrt{x-2}} \qquad \text{or} \qquad y^2 &=& x-2 \\ \mathbf{ x } &=& \mathbf{y^2+2} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \sqrt{x+14-8\sqrt{x-2}} + \sqrt{x+23-10\sqrt{x-2}} &=& 3 \\\\ \sqrt{y^2+2+14-8y} + \sqrt{y^2+2+23-10y} &=& 3 \\\\ \sqrt{y^2-8y+16} + \sqrt{y^2-10y+25} &=& 3 \\\\ && \boxed{ y^2-8y+16 = (y-4)^2=(4-y)^2} \\ && \boxed{ y^2-10y+25 = (y-5)^2=(5-y)^2} \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline \text{Case }1: \\ & y^2-8y+16 = (y-4)^2 \\ & y^2-10y+25 = (y-5)^2 \\ \hline & \sqrt{(y-4)^2} + \sqrt{(y-5)^2 } &=& 3 \\\\ & y-4 + y-5 &=& 3 \\\\ & 2y-9 &=& 3 \\\\ & 2y &=& 9+3 \\\\ & 2y &=& 12 \\\\ & \mathbf{y} &=& \mathbf{6} \\ \hline \end{array} \begin{array}{|lrcll|} \hline \text{Case }2: \\ & y^2-8y+16 = (y-4)^2 \\ & y^2-10y+25 = (5-y)^2 \\ \hline & \sqrt{(y-4)^2} + \sqrt{(5-y)^2 } &=& 3 \\\\ & y-4 + 5-y &=& 3 \\\\ & \mathbf{1} &\ne& \mathbf{3} \qquad \text{no solution!} \\ \hline \end{array}\\ \begin{array}{|lrcll|} \hline \text{Case }3: \\ & y^2-8y+16 = (4-y)^2 \\ & y^2-10y+25 = (5-y)^2 \\ \hline & \sqrt{(4-y)^2} + \sqrt{(5-y)^2 } &=& 3 \\\\ & 4-y + 5-y &=& 3 \\\\ & -2y+9 &=& 3 \\\\ & 2y &=& 9-3 \\\\ & 2y &=& 6 \\\\ & \mathbf{y} &=& \mathbf{3} \\ \hline \end{array} \begin{array}{|lrcll|} \hline \text{Case }4: \\ & y^2-8y+16 = (4-y)^2 \\ & y^2-10y+25 = (y-5)^2 \\ \hline & \sqrt{(4-y)^2} + \sqrt{(y-5)^2 } &=& 3 \\\\ & 4-y + y-5 &=& 3 \\\\ & \mathbf{-1} &\ne& \mathbf{3} \qquad \text{no solution!} \\ \hline \end{array}\\$

$\begin{array}{|rcll|} \hline \text{First solution}: \\ y= 6 \\ \hline x &=& y^2+2 \\ x &=& 6^2+2 \\ \mathbf{x} &=& \mathbf{38} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{Second solution}: \\ y= 3 \\ \hline x &=& y^2+2 \\ x &=& 3^2+2 \\ \mathbf{x} &=& \mathbf{11} \\ \hline \end{array}$