$\sqrt{x-1}$ = $x-4$
⇔ $(\sqrt{x-1})^{2}$ = $(x-4)^{2}$
⇔ $x-1$ = $x^{2} - 8x + 16$
⇔ $x^{2} - 9x +17 = 0$
⇔ $Δ = b^{2} - 4ac = (-9)^{2} - 4.1.17 = 13$
⇔ $\sqrt{Δ} = \sqrt{13}$
⇔ $x_{1} = \frac{-b-\sqrt{Δ}}{2a} = \frac{9-\sqrt{13}}{2}$
$x_{2} = \frac{-b+\sqrt{Δ}}{2a} = \frac{-9-\sqrt{13}}{2}$