To solve the quadratic equation \( x^2 + 3x - 8 = 0 \) using the quadratic formula, we use the formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, the coefficients are:
- \( a = 1 \)
- \( b = 3 \)
- \( c = -8 \)
Now we can compute the discriminant \( b^2 - 4ac \):
\[ b^2 = 3^2 = 9 \] \[ 4ac = 4 \cdot 1 \cdot (-8) = -32 \] \[ b^2 - 4ac = 9 - (-32) = 9 + 32 = 41 \]
Now, substitute \( a \), \( b \), and the discriminant back into the formula:
\[ x = \frac{-3 \pm \sqrt{41}}{2 \cdot 1} \]
This simplifies to:
\[ x = \frac{-3 \pm \sqrt{41}}{2} \]
Thus, the two solutions are:
\[ x = \frac{-3 + \sqrt{41}}{2}, \quad x = \frac{-3 - \sqrt{41}}{2} \]
In conclusion, your answer is:
\[ x = \frac{-3 + \sqrt{41}}{2}, \frac{-3 - \sqrt{41}}{2} \]
You can enter this as:
\((-3 + \sqrt{41})/2, (-3 - \sqrt{41})/2\)