Question 1178991

 2𝑥^2 + (1 − 2𝑛)𝑥 − 𝑛 = 0 

comparing the equation with ax^2 +bx +c

a= 2,  b=(1-2n), c=-n

b^2 = (1-2n)^2 
1-4n +4n^2

4ac = (4)(2)(-n)
= -8n
b^2 -4ac
1-4n +4n^2 -(-8n)
4n^2 -3n +1
4n^2+4n+1
(2n+1)^2
For all values of n, (2n+1)^2 >0
Therefore roots are real and unequal