To solve the quadratic equation \( x^2 + 3x - 5 = 0 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this equation, the coefficients are:
- \( a = 1 \) (the coefficient of \( x^2 \))
- \( b = 3 \) (the coefficient of \( x \))
- \( c = -5 \) (the constant term)
Now, we can substitute these values into the quadratic formula.
- Compute the discriminant:
\[ b^2 - 4ac = 3^2 - 4(1)(-5) = 9 + 20 = 29 \]
- Substitute into the quadratic formula:
\[ x = \frac{-3 \pm \sqrt{29}}{2 \times 1} = \frac{-3 \pm \sqrt{29}}{2} \]
This gives us the two solutions:
\[ x_1 = \frac{-3 + \sqrt{29}}{2} \] \[ x_2 = \frac{-3 - \sqrt{29}}{2} \]
So, the solutions to the equation \( x^2 + 3x - 5 = 0 \) are:
\[ x = \frac{-3 + \sqrt{29}}{2} \quad \text{and} \quad x = \frac{-3 - \sqrt{29}}{2} \]