Dividing by e^2x

$${e^{2x}} + 4{e^x} - 58 + 4{e^{ - x}} + {e^{ - 2x}} = 0$$

$$ \Rightarrow {({e^x} + {e^{ - x}})^2} + 4({e^x} + {e^{ - x}}) - 60 = 0$$

Let $${e^x} + {e^{ - x}} = t \in [2,\infty )$$

$$ \Rightarrow {t^2} + 4t - 60 = 0$$

$$ \Rightarrow t = 6$$ is only possible solution

$${e^x} + {e^{ - x}} = 6 \Rightarrow {e^{2x}} - 6{e^x} + 1 = 0$$

Let $${e^x} = p$$,

$${p^2} - 6p + 1 = 0$$

$$ \Rightarrow p = {{3 + \sqrt 5 } \over 2}$$ or $${{3 - \sqrt 5 } \over 2}$$

So $$x = \ln \left( {{{3 + \sqrt 5 } \over 2}} \right)$$ or $$\ln \left( {{{3 - \sqrt 5 } \over 2}} \right)$$