Use the discriminant to determine how many real number solutions exist for the quadratic equation –4j2 + 3j – 28 = 0.

- 4j² + 3j - 28 =0
It's quadratic equation of the form ax² + bx + c, where a = -4, b=3 and c =- 28.

The discriminent Δ = b² - 4.a.c
Δ = 3² - 4(-4)(-28)
Δ = -439 

If Δ ≥ 0 there are either 2 roots or one double if = 0)
if Δ< 0, there are no real roots (but 2 imaginary roots)
Since Δ < 0, then there is no real roots

The answer is 10 bud