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
