Can you please solve the equation (1/(b+3))+2 = (b^2-3)/(b^2+12b+27)

topic: solve rational equations. the dashes "/" represent fraction bars

(1 / (b + 3)) + 2 = (b^2 - 3) / (b^2 + 12b + 27)
((b + 9) / ((b + 3)(b + 9))) + ((2(b + 3)(b + 9)) / ((b + 3)(b + 9)) = (b^2 - 3) / ((b + 3)(b + 9))
((b + 9) / ((b + 3)(b + 9))) + ((2b^2 + 24b + 54) / ((b + 3)(b + 9)) = (b^2 - 3) / ((b + 3)(b + 9))
((b + 9) + (2b^2 + 24b + 54)) / ((b + 3)(b + 9)) = (b^2 - 3) / ((b + 3)(b + 9))
((b + 9) + (2b^2 + 24b + 54)) = (b^2 - 3)
2b^2 + 25b + 63 = b^2 - 3
2b^2 - b^2 + 25b + 63 + 3 = 0
b^2 + 25b + 66 = 0
b^2 + 22b + 3b + 66 = 0
(b^2 + 22b) + (3b + 66) = 0
b(b + 22) + 3(b + 22) = 0
(b + 3)(b + 22) = 0
b + 3 = 0 or b + 22 = 0
b = -3 or b = -22

Notice that b + 3 is in the denominator of (1 / (b + 3)).
Thus, if we substitute b = -3 in, we will have 1 / 0, which is undefined.
Thus, we must reject b = -3.
Hence, b = -22 is the only answer.