algebra 2 help?

Question

Solve: log2(x) + log2(x + 2) = 3

Answer 1

log2(x) + log2(x + 2) = 3
log2[x^2 + 2x] = log2 8
x^2 + 2x - 8 = 0
(x-2)(x+4) = 0
x = 2 or x = -4
x= -4 rejected because x must be > 0 for logarithms to be defined.

Answer 2

log[2] x + log[2] (x+3) = 3
log[2] x(x+3) = 3
x(x+3) = 2^3 = 8
x² + 3x - 8 = 0
x = -3/2 ± [√(9 + 32)]/2
x = -3/2 ± (√41)/2
can't use neg solution since log neg is undefined, so
x = -1.5 + 3.2016
x = 1.70156