To solve the quadratic equation \( 12k^2 + 37k + 3 = 0 \), we can use the quadratic formula:
\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 12 \), \( b = 37 \), and \( c = 3 \).
First, we need to compute the discriminant \( b^2 - 4ac \):
\[ b^2 = 37^2 = 1369 \]
\[ 4ac = 4 \cdot 12 \cdot 3 = 144 \]
Now, subtract to find the discriminant:
\[ b^2 - 4ac = 1369 - 144 = 1225 \]
Next, we compute the square root of the discriminant:
\[ \sqrt{1225} = 35 \]
Now, we can plug the values into the quadratic formula:
\[ k = \frac{-37 \pm 35}{2 \cdot 12} \]
This gives us two potential solutions:
- For \( k = \frac{-37 + 35}{24} \):
\[ k = \frac{-2}{24} = \frac{-1}{12} \]
- For \( k = \frac{-37 - 35}{24} \):
\[ k = \frac{-72}{24} = -3 \]
Thus, the solutions for \( k \) are:
\[ \boxed{-3, -\frac{1}{12}} \]