Найдём координаты вектора \(\overrightarrow {AB} \): \(\overrightarrow {AB} = \left( {6-2x;4x + 2} \right).\)
Длина вектора \(\vec d\left( {x;y} \right)\) равна: \(\left| {\vec d} \right| = \sqrt {{x^2} + {y^2}} \).
Тогда: \(\left| {\overrightarrow {AB} } \right| = \sqrt {{{\left( {6-2x} \right)}^2} + {{\left( {4x + 2} \right)}^2}} .\)
Следовательно: \(\sqrt {{{\left( {6-2x} \right)}^2} + {{\left( {4x + 2} \right)}^2}} = 14\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,36-24x + 4{x^2} + 16{x^2} + 16x + 4\, = 196\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,5{x^2}-2x-39\, = 0\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 3,\,\,\,\,\,\,\,\,}\\{x = -2,6.}\end{array}} \right.\)
Так как \(x < 0\), то \(x = -2,6\).
Ответ: \(-2,6\).