To solve the equation \(4x^2 + 12x = 135\), we first rearrange it into standard quadratic form:
\[ 4x^2 + 12x - 135 = 0 \]
Next, we can use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation, the coefficients are:
- \(a = 4\)
- \(b = 12\)
- \(c = -135\)
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 substitute back into the quadratic formula:
\[ x = \frac{-12 \pm \sqrt{2304}}{2 \cdot 4} \]
Calculating \(\sqrt{2304}\):
\[ \sqrt{2304} = 48 \]
Now substituting that into the quadratic formula:
\[ x = \frac{-12 \pm 48}{8} \]
This gives us two solutions:
-
Positive solution: \[ x = \frac{-12 + 48}{8} = \frac{36}{8} = 4.5 \]
-
Negative solution: \[ x = \frac{-12 - 48}{8} = \frac{-60}{8} = -7.5 \]
The positive solution is:
\[ \boxed{4.5} \]