We have \( \Large |x^{2}+4x+3|+2x+5=0 \)

have two cases arise

Case l: When \( \Large x^{2}+4x+3+2x+5=0 \)

=> \( \Large x^{2}+6x+8=0 \)

=> \( \Large \left(x+2\right) \left(x+4\right)=0 \)

=> \( \Large x = -2, -4 \)

x = -2 is not satisfying the condition \( \Large x^{2}+4x+3>0 \),

so. x = -4 is the only solution of given equation.

Case II: When \( \Large x^{2}+4x+3<0 \)

This gives \( \Large - \left(x^{2}+4x+3\right)+2x+5=0 \)

=> \( \Large -x^{2}-2x+2=0 \)

=> \( \Large x^{2}+2x-2=0 \)

=> \( \Large \left(x+1+\sqrt{3}\right) \left(x+1-\sqrt{3}\right)=0 \)

=> \( \Large x = -1+\sqrt{3}, -1-\sqrt{3} \)

Hence, \( \Large x=- \left(1+\sqrt{3}\right) \) satisfy the given condition.

Since, \( \Large x^{2}+4x+3<0 \) while \( \Large x= -1+\sqrt{3} \) is not satisfying the condition. Thus number of real solutions are two.