To determine the number of real solutions to the quadratic equation \(4x^2 + x + 3 = 0\), we can use the discriminant. For a quadratic equation of the form \(ax^2 + bx + c = 0\), the discriminant \(D\) is given by:
\[ D = b^2 - 4ac \]
In this case, we have:
- \(a = 4\)
- \(b = 1\)
- \(c = 3\)
Let's calculate the discriminant:
\[ D = 1^2 - 4 \cdot 4 \cdot 3 \]
Calculating \(1^2\):
\[ 1^2 = 1 \]
Calculating \(4 \cdot 4 \cdot 3\):
\[ 4 \cdot 4 \cdot 3 = 48 \]
Now substituting these values into the discriminant formula:
\[ D = 1 - 48 = -47 \]
The discriminant \(D\) is negative (\(-47 < 0\)). This indicates that the quadratic equation has no real solutions.
Thus, the conclusion is:
\[ \text{The quadratic equation } 4x^2 + x + 3 = 0 \text{ has 0 real solutions.} \]