we can rearrange the equation as x^2+2*x*1+1^2=0

now, as we may know that (a+b)^2=a^2+2ab+b^2

so the above equation becomes (x+1)^2=0

now taking square root on both sides

x+1=0

subtracting 1 from both sides

x+1-1=0-1

or, x=-1

so the solution is x=-1

we can check the number of solutions of equations in form ax^2 + b x+ c

by checking the discriminant= b^2 - 4 ac

here x^2 + 2 x +1 = 0

discriminant= b^2 - 4 ac

=2^2 - 4*1*1

=0

if discriminant = 0 it will have only one real solution...................

x^2 + 2 x +1 = 0

x^2 + x+x + 1=0

x(x+1) + 1(x+1) = 0

(x+1) (x+1) = 0

x + 1 = 0

x= -1 ................................................................ answer

hope you understood...please message if you have any doubts...thank you

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