how many real number solutions does the equation have
y = -4x^2 + 7x - 8
no solutions (is it this one)
one solution
two solutions
infinitely many solutions
you mean how many zeros I think
look at b^2-4ac
49 -16*8 is NEGATIVE
so two complex conjugate solutions, NO real solutions
To determine the number of real number solutions of an equation, we need to analyze the discriminant of the quadratic equation. The general form of a quadratic equation is ax^2 + bx + c = 0, where a, b, and c are constants.
In this case, the equation is y = -4x^2 + 7x - 8, which is in the form of y = ax^2 + bx + c. To find the discriminant, we can use the formula: D = b^2 - 4ac.
Comparing it with the given equation:
a = -4,
b = 7,
c = -8.
Now, substituting these values into the formula for the discriminant:
D = (7)^2 - 4(-4)(-8)
D = 49 - 128
D = -79
The discriminant is -79. Since the discriminant is negative, it means that the quadratic equation has no real number solutions.
So, the correct answer is: no solutions.