To determine how many real solutions the quadratic equation \(4x^2 + x + 3 = 0\) has, we can use the discriminant from the quadratic formula:
The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a\), \(b\), and \(c\) are the coefficients from the equation \(ax^2 + bx + c = 0\).
For the equation \(4x^2 + x + 3 = 0\):
- \(a = 4\)
- \(b = 1\)
- \(c = 3\)
Now, we calculate the discriminant \(D = b^2 - 4ac\):
\[ D = (1)^2 - 4(4)(3) = 1 - 48 = -47 \]
Since the discriminant \(D\) is negative (\(-47 < 0\)), this means the quadratic equation has no real solutions.
Therefore, the correct answer is:
- no real solutions
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