Answer:
[tex] \boxed{\sf d = -1 \ \ \ or \ \ \ d = -7} [/tex]
Step-by-step explanation:
[tex] \sf Solve \: for \: d \: over \: the \: real \: numbers: \\ \sf \implies {(d + 4)}^{2} = 9 \\ \\ \sf Take \: the \: square \: root \: of \: both \: sides: \\ \sf \implies \sqrt{ {(d + 4)}^{2} } = \sqrt{9} \\ \\ \sf 9 = {3}^{2} : \\ \sf \implies \sqrt{ {(d + 4)}^{2} } = \sqrt{ {(3)}^{2} } \\ \\ \sf \implies d + 4 = \pm 3 \\ \\ \sf \implies d + 4 = 3 \: \: \: \: \: or \: \: \: \: \: d + 4 = - 3 \\ \\ \sf Subtract \: 4 \: from \: both \: sides: \\ \sf \implies d + (4 - 4)= 3 - 4 \: \: \: \: \: or \: \: \: \: \: d + (4 - 4) = - 3 - 4 \\ \\ \sf \implies d = - 1 \: \: \: \: \: or \: \: \: \: \: d = - 7[/tex]
