To find the integer solutions of the quadratic equation \(4x^2 + 11x - 20 = 0\), we can use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the equation \(4x^2 + 11x - 20 = 0\), we identify \(a = 4\), \(b = 11\), and \(c = -20\).
First, we need to calculate the discriminant \(D\):
\[ D = b^2 - 4ac \]
Substituting the values of \(a\), \(b\), and \(c\):
\[ D = 11^2 - 4 \times 4 \times (-20) \] \[ D = 121 + 320 \] \[ D = 441 \]
Next, we compute the square root of the discriminant:
\[ \sqrt{D} = \sqrt{441} = 21 \]
Now we can apply the quadratic formula:
\[ x = \frac{-11 \pm 21}{2 \times 4} \]
This gives us two potential solutions:
- For the positive case:
\[ x = \frac{-11 + 21}{8} = \frac{10}{8} = \frac{5}{4} \]
- For the negative case:
\[ x = \frac{-11 - 21}{8} = \frac{-32}{8} = -4 \]
Thus, the two solutions to the equation \(4x^2 + 11x - 20 = 0\) are:
\[ x = \frac{5}{4} \quad \text{and} \quad x = -4 \]
Among these solutions, the only integer solution is:
\(-4\)
Therefore, the number of integer solutions is:
\[ \boxed{1} \]