Hi there! The numbers can be either -15 or 3.
Let the number be represented by X.
[tex](x + 6) {}^{2} = 81[/tex]
First we work out the parentheses.
[tex] {x}^{2} + 12x + 36 = 81[/tex]
Now subtract 81 from both sides.
[tex] {x}^{2} + 12x - 45 = 0[/tex]
Using the sum of 12 and a product of 45 we get.
[tex](x + 15)(x - 3) = 0[/tex]
AB = 0 gives A = 0 or B = 0
[tex]x + 15 = 0 \: \: or \: \: x - 3 = 0[/tex]
[tex]x = - 15 \: \: or \: \: x = 3[/tex]
Therefore, the numbers are either -15 or 3.
