To determine the number of integer solutions for the equation \(4x^2 + 11x - 20 = 0\), we start by applying the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our case, the coefficients are:
- \(a = 4\)
- \(b = 11\)
- \(c = -20\)
First, we need to calculate the discriminant \(b^2 - 4ac\):
\[ b^2 = 11^2 = 121 \] \[ 4ac = 4 \cdot 4 \cdot (-20) = -320 \]
Now, adding these together:
\[ b^2 - 4ac = 121 - (-320) = 121 + 320 = 441 \]
Next, we check if the discriminant is a perfect square, which will indicate whether the solutions are rational (and possibly integers). We find the square root of \(441\):
\[ \sqrt{441} = 21 \]
Now we substitute back into the quadratic formula:
\[ x = \frac{-11 \pm 21}{2 \cdot 4} \]
This results in two possible values for \(x\):
- For the positive case:
\[ x = \frac{-11 + 21}{8} = \frac{10}{8} = \frac{5}{4} \]
- For the negative case:
\[ x = \frac{-11 - 21}{8} = \frac{-32}{8} = -4 \]
Now we examine the results:
- The first solution \(x = \frac{5}{4}\) is not an integer.
- The second solution \(x = -4\) is an integer.
Thus, among the two solutions of the equation \(4x^2 + 11x - 20 = 0\), only one solution is an integer.
Therefore, the number of integer solutions is:
\[ \boxed{1} \]