Question

Find all solutions for x^2 + 3x - 18 = 0

Answer 1

To solve the quadratic equation \( x^2 + 3x - 18 = 0 \), we can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the given equation, the coefficients are:
- \( a = 1 \)
- \( b = 3 \)
- \( c = -18 \)

Now, we can calculate the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot (-18) = 9 + 72 = 81
\]

Since the discriminant is positive, there will be two distinct real solutions. Now we can substitute the values into the quadratic formula:

\[
x = \frac{-3 \pm \sqrt{81}}{2 \cdot 1}
\]

Calculating the square root:

\[
\sqrt{81} = 9
\]

Now plug it back into the formula:

\[
x = \frac{-3 \pm 9}{2}
\]

This gives us two solutions:

1. For the positive case:

\[
x = \frac{-3 + 9}{2} = \frac{6}{2} = 3
\]

2. For the negative case:

\[
x = \frac{-3 - 9}{2} = \frac{-12}{2} = -6
\]

Thus, the solutions to the equation \( x^2 + 3x - 18 = 0 \) are:

\[
\boxed{3} \quad \text{and} \quad \boxed{-6}
\]