how many real solutions does 4x^2+x+3=0 have

Answer 1

The number of real solutions for a quadratic equation can be determined by the discriminant (D) of the equation, which is given by D = b^2 - 4ac.

For the equation 4x^2 + x + 3 = 0, with a = 4, b = 1, and c = 3, the discriminant can be calculated as follows:

D = (1)^2 - 4(4)(3) = 1 - 48 = -47

Since the discriminant (-47) is negative, there are no real solutions to the equation.