`5`$(x-4)^{2}$ `- 7 = 13`
`⇔ 5`$(x-4)^{2}$ ` = 13+ 7`
`⇔ 5`$(x-4)^{2}$ ` = 20`
`⇔`$(x-4)^{2}$ ` = 20 : 5`
`⇔`$(x-4)^{2}$ ` = 4`
`⇔` $(x-4)^{2}$ ` = (±2)²`
`⇔` \(\left[ \begin{array}{l}x - 4=2\\x - 4=-2\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x =6\\x =2\end{array} \right.\)
Vậy `x ∈ {6; -2}`