`\text{a) ( x+2)² - 9 = 0}`
`\text{(x+2)² - 3² = 0}`
`\text{(x+2 - 3 )( x+ 2+3 ) = 0}`
`\text{( x-1 ) ( x +5 ) = 0 }`
⇔\(\left[ \begin{array}{l}x-1 = 0\\x+5 = 0 \end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x = 1\\x = -5 \end{array} \right.\)
Vậy x = 1 hay x = - 5