Given equation $e^{4x}+4e^{3x}-58e^{2x}+4e^x+1=0$

Let $f\left(x\right)=e^{2x}\left(e^{2x}+\frac{1}{e^{2x}}+4\left(e^x+\frac{1}{e^x}\right)-58\right)$

Now let $e^x+\frac{1}{e^x}=t$

i.e. $t^2+4t-58=0$

$t=\frac{-4\pm \sqrt{16+4\times 58}}{2}$

$=\frac{-4\pm 2\sqrt{62}}{2}$

$t=-2+2\sqrt{62}$ is possible and $t=-2-2\sqrt{62}$ is not possible as $t\ge 2$

i.e. ${\mathrm{e}}^x+\frac{1}{{\mathrm{e}}^x}=-2+2\sqrt{62}$

${\mathrm{e}}^{2x}-\left(-2+2\sqrt{62}\right){\mathrm{e}}^x+1=0$

For real solution, discriminant $248-8\sqrt{62}>0$

Also $\frac{-b}{2a}=\frac{-2+2\sqrt{62}}{2}=\sqrt{62}-1>0$

Hence, both roots of the equation are positive real roots.