Answer
Sample answer.
$x^{2}+2x=0$ (has 2 distinct solutions)
$x^{2}+2x+2=0$ (has no real solutions)
$x^{2}+2x+1=0$ (has one real, repeated solution)
Work Step by Step
Sample answer
Take $(x+1)^{2}=0$
It has one solution, $-1$.
In standard form, the equation is $x^{2}+2x+1=0$
Next, $(x+1)^{2}-1^{2}=0$
has a difference of squares on the LHS, which can be factored
$(x+1-1)(x+1+1)=0$
$x(x+2)=0$, and it has 2 real solutions.
In standard form, $x^{2}+2x=0$
Finally, $(x+1)^{2}+1^{2}=0$
has no real solutions, because $(x+1)^{2}$ can not be $-1.$
In standard form, $x^{2}+2x+2=0$
