Use the discriminant to determine how many real number solutions exist for the quadratic equations -4j+3j-28=0

The problem corrected is the following

4j^2+3j-28=0

In a quadratic equation

q(x) = ax^2 + bx + c 

The discriminant is = b^2 - 4ac

If b^2 - 4ac > 0, then the roots are real.

If b^2 - 4ac < 0 then the roots are imaginary

In this problem

b^2 - 4ac = 3^2 – 4(4)(-28)=457

457>0  then the two roots must be real

For the quadratic equation there are two real solutions

its the fourth option

okay wait before i answer, is this for your points on a graph?