To determine how many solutions of the quadratic equation \( 4x^2 + 11x - 20 = 0 \) are integers, we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 4 \), \( b = 11 \), and \( c = -20 \).
First, we calculate the discriminant \( D \):
\[ D = b^2 - 4ac = 11^2 - 4(4)(-20) \]
Calculating \( 11^2 \):
\[ 11^2 = 121 \]
Now calculating \( 4(4)(-20) \):
\[ 4 \cdot 4 \cdot -20 = -320 \]
So, we find \( D \):
\[ D = 121 - (-320) = 121 + 320 = 441 \]
Next, we check if \( D \) is a perfect square. The square root of \( 441 \) is:
\[ \sqrt{441} = 21 \]
Now substituting \( D \) back into the quadratic formula:
\[ x = \frac{-11 \pm 21}{2 \cdot 4} = \frac{-11 \pm 21}{8} \]
This gives us two possible solutions:
\[ x_1 = \frac{-11 + 21}{8} = \frac{10}{8} = \frac{5}{4} \]
\[ x_2 = \frac{-11 - 21}{8} = \frac{-32}{8} = -4 \]
Now, we have the two solutions \( x_1 = \frac{5}{4} \) and \( x_2 = -4 \).
Out of these, only \( x_2 = -4 \) is an integer.
Thus, the number of integer solutions for the quadratic equation \( 4x^2 + 11x - 20 = 0 \) is:
\[ \boxed{1} \]