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

The quadratic formula, for the equation   ax² + bx + c = 0,   where   a ≠ 0, is:
x = (-b plus or minus √(b² - 4ac)) ÷ 2a. From this formula...
...the discriminant of  ax² + bx + c = 0  is given by   Δ = b² - 4ac.

If Δ > 0, then there are two real roots. If Δ < 0, there are no real solutions.
If Δ = 0, there is exactly 1 real root.

Using this, we can substitute the coefficients for j into the discriminant formula. Δ = 3² - 4(-4)(-28), which equals 9 - (448), which is clearly less than 0.

So there are NO REAL ROOTS to   -4j² + 3j - 28 = 0

Hope this helps!

The answer would be d i think but i’m for sure it is not a and c