The number of solutions of $x^2+2x+1=0$  is

${\left(x+1\right)}^2=0$

$\left(x+1\right)\left(x+1\right)=0$

$x=-1orx=-1$

$\left(or\right)$

$a=1,b=2,c=1$

$\Delta =b^2-4ac$

$={\left(2\right)}^2-4\left(1\right)\left(1\right)$

$=4-4$

$=0$

It has one solution number of solution is one (repeated values)