To solve the quadratic equation \(4x^2 + 4x - 15 = 0\) using the quadratic formula, we start with the standard form of a quadratic equation:
\[ ax^2 + bx + c = 0 \]
Here, \(a = 4\), \(b = 4\), and \(c = -15\).
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
First, we calculate the discriminant \(b^2 - 4ac\):
\[ b^2 = 4^2 = 16 \] \[ 4ac = 4 \cdot 4 \cdot (-15) = -240 \] \[ b^2 - 4ac = 16 + 240 = 256 \]
Now we can substitute these values into the quadratic formula:
\[ x = \frac{-4 \pm \sqrt{256}}{2 \cdot 4} \]
Calculating \(\sqrt{256}\):
\[ \sqrt{256} = 16 \]
Now plug this back into the formula:
\[ x = \frac{-4 \pm 16}{8} \]
This gives us two possible solutions:
- Using the plus sign:
\[ x = \frac{-4 + 16}{8} = \frac{12}{8} = \frac{3}{2} = 1.5 \]
- Using the minus sign:
\[ x = \frac{-4 - 16}{8} = \frac{-20}{8} = -\frac{5}{2} = -2.5 \]
Thus, the solutions to the equation \(4x^2 + 4x - 15 = 0\) are:
\[ x = 1.5 \quad \text{and} \quad x = -2.5 \]
The correct response is: \(x = 1.5\), \(x = -2.5\).