Let
x--------> the length side of the original square paper
we know that
the area of the original square paper is equal to
[tex] A1=x*x=x^{2} \ units^{2} [/tex]
the area of the remaining piece of paper is equal to
[tex] A2=x^{2} -2x\\ A2=120\ units^{2} \\ so\\ x^{2} -2x=120\\ x^{2} -2x-120=0 [/tex]
therefore
the answer is the option
x2 − 2x − 120 = 0