Answer:
[tex]\textbf{$x$ can take two values $1$ or $-9$.}\\[/tex]
Step-by-step explanation:
[tex]\textup{Given, ${(x + 4)}^2 = 25$}\\\textup{If we take square root on both sides, then we would have:}\\$ {(x+4)} = \pm \sqrt{25} = \pm 5 $\\$ \implies x + 4 = 5 \hspace{15mm} $ or \hspace{15mm} $ x + 4 = -5 $\\$ \implies x = 1 \hspace{22mm} $ or \hspace{15mm} $x = -9 $ \\\textup{Verify your answer by expanding ${(x+4)}^2 = 25$}\\\textup{Solving this we get}\\$ x^2 + 8x - 9 = 0 $\\\textup{which again yields}\\$ x = -9, 1 $[/tex]
