To solve the equation \(4x^2 + 12x = 135\), we first rearrange it into standard form:
\[ 4x^2 + 12x - 135 = 0 \]
Next, we can apply the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 4\), \(b = 12\), and \(c = -135\).
Now, we calculate the discriminant \(b^2 - 4ac\):
\[ b^2 = 12^2 = 144 \] \[ 4ac = 4 \cdot 4 \cdot (-135) = -2160 \] \[ b^2 - 4ac = 144 + 2160 = 2304 \]
Now, we take the square root of the discriminant:
\[ \sqrt{2304} = 48 \]
Now substituting back into the quadratic formula:
\[ x = \frac{-12 \pm 48}{2 \cdot 4} = \frac{-12 \pm 48}{8} \]
Calculating the two possible solutions:
- For \(x = \frac{-12 + 48}{8}\):
\[ x = \frac{36}{8} = \frac{9}{2} = 4.5 \]
- For \(x = \frac{-12 - 48}{8}\):
\[ x = \frac{-60}{8} = -7.5 \]
Since we are looking for the positive solution, we take:
\[ \boxed{4.5} \]