Log2(x^2+3)=1+log2(x+3)
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ОДЗ:
{x^2+3>0; x e R
{x+3>0; x>-3
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log2(x^2+3) = log2(2)+ log2(x+3)
log2(x^2+3)=log2(2x+6)
x^2+3=2x+6
x^2+3-2x-6=0
x^2-2x-3=0
D=(-2)^2-4*1*(-3)=16
x1=(2-4)/2=-1
x2=(2+4)/2=3
Ответ: {-1;3}