A Point in Relation to the Vertices of an Equilateral Triangle

Possible sides of the triangle: $\approx \quad \underline { 8 } .5363...\quad \& \quad \underline { 2 } .0324...\\ Sum\quad of\quad unit's\quad digits:\quad 8\quad +\quad 2\quad =\quad \boxed { 10 }$

Actual solution: $3\left( { a }^{ 4 }\quad +\quad { b }^{ 4 }\quad +\quad { c }^{ 4 }\quad +\quad { x }^{ 4 } \right) \quad =\quad { \left( { a }^{ 2 }\quad +\quad { b }^{ 2 }\quad +\quad { c }^{ 2 }\quad +\quad { x }^{ 2 } \right) }^{ 2 }$ where $a$ , $b$ , and $c$ are the distances from the vertices, and $x$ is the length of the sides of the triangle.

$\Longrightarrow \quad 3\left( 4^{ 4 }\quad +\quad { 5 }^{ 4 }\quad +\quad { 6 }^{ 4 }\quad +\quad { x }^{ 4 } \right) \quad =\quad { \left( { 4 }^{ 2 }\quad +\quad { 5 }^{ 2 }\quad +\quad { 6 }^{ 2 }\quad +\quad { x }^{ 2 } \right) }^{ 2 }\\ \Longrightarrow \quad 3{ x }^{ 4 }\quad +\quad 6531\quad =\quad { x }^{ 4 }\quad +\quad 154{ x }^{ 2 }\quad +\quad 5929\\ \Longrightarrow \quad { x }^{ 4 }\quad -\quad 77{ x }^{ 2 }\quad +\quad 301\quad =\quad 0\\ \Longrightarrow \quad { x }^{ 2 }\quad =\quad \frac { 77\quad \pm \quad \sqrt { { 77 }^{ 2 }\quad -\quad 4\left( 301 \right) } }{ 2 } \\ \Longrightarrow \quad x\quad =\quad \sqrt { \frac { 77\quad \pm \quad 15\sqrt { 21 } }{ 2 } } \quad \because \quad x\quad >\quad 0\\ \Longrightarrow \quad x\quad \approx \quad 8.5363...\quad and\quad 2.0324$