Question

solutions to the equation \(x^{2}+3x-5=0\) are \(x=\frac{-3+\sqrt{29}}{2}\) and \(x=\frac{-3-\sqrt{29}}{2}\).

Answer 1

The equation you've given, \(x^{2}+3x-5=0\), can be solved using the quadratic formula:

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

In this case, the coefficients are:
- \(a = 1\)
- \(b = 3\)
- \(c = -5\)

Plugging these values into the quadratic formula:

1. Calculate \(b^2 - 4ac\):

\[
b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot (-5) = 9 + 20 = 29
\]

2. Now, substitute back into the quadratic formula:

\[
x = \frac{-3 \pm \sqrt{29}}{2 \cdot 1} = \frac{-3 \pm \sqrt{29}}{2}
\]

Thus, the solutions to the equation \(x^{2}+3x-5=0\) are indeed:

\[
x = \frac{-3+\sqrt{29}}{2} \quad \text{and} \quad x = \frac{-3-\sqrt{29}}{2}
\]

Your statement is correct!