Determine the number and type of solutions for the quadratic equation:

4X^2 + 3X= -8

4 x^2 + 3 x + 8 = 0

the nature of the discriminant determines the type of solutions

b^2 - 4 a c = 9 - 32 = -23 ... negative and not a perfect square
... the solutions are imaginary , two complex conjugates

4 x^2 + 3x +8 = 0

b^2-4ac = 9 - 128
Negative so solutions are complex

Damon got the right value for the discriminant

final result doesn't change

To determine the number and type of solutions for a quadratic equation, we can consider its discriminant. The discriminant is the expression found under the square root sign of the quadratic formula. For a quadratic equation in the form of ax^2 + bx + c = 0, the discriminant is given as:

Δ = b^2 - 4ac

If the discriminant is greater than zero (Δ > 0), the quadratic equation has two distinct real solutions. If the discriminant is equal to zero (Δ = 0), the quadratic equation has one real solution (a double root). If the discriminant is less than zero (Δ < 0), the quadratic equation has no real solutions and instead has two complex solutions.

Let's apply this to the given equation: 4X^2 + 3X = -8

First, we can rewrite the equation in the standard form ax^2 + bx + c = 0:
4X^2 + 3X + 8 = 0

Next, we can identify the values of a, b, and c:
a = 4
b = 3
c = 8

Now, let's calculate the discriminant (Δ) for the given equation:
Δ = b^2 - 4ac
= (3)^2 - 4(4)(8)
= 9 - 128
= -119

Since the discriminant (Δ) is less than zero (Δ < 0), the quadratic equation 4X^2 + 3X + 8 = 0 has no real solutions.