To solve the quadratic equation \( x^2 + 5x - 36 = 0 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this equation:
- \( a = 1 \)
- \( b = 5 \)
- \( c = -36 \)
Now, we substitute the values into the formula:
- Calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 5^2 = 25 \]
\[ 4ac = 4 \cdot 1 \cdot (-36) = -144 \]
So,
\[ b^2 - 4ac = 25 + 144 = 169 \]
- Now substitute back into the quadratic formula:
\[ x = \frac{-5 \pm \sqrt{169}}{2 \cdot 1} \]
Since \( \sqrt{169} = 13 \), we have:
\[ x = \frac{-5 \pm 13}{2} \]
This gives us two potential solutions:
- \( x = \frac{-5 + 13}{2} = \frac{8}{2} = 4 \)
- \( x = \frac{-5 - 13}{2} = \frac{-18}{2} = -9 \)
The positive solution is:
\[ \boxed{4} \]
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