To solve the quadratic equation \( x^2 + 3x - 5 = 0 \), we can use the quadratic formula, which is given by:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]
where \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation in the standard form \( ax^2 + bx + c = 0 \).
For the equation \( x^2 + 3x - 5 = 0 \):
- \( a = 1 \)
- \( b = 3 \)
- \( c = -5 \)
Now, let's substitute the values of \( a \), \( b \), and \( c \) into the quadratic formula:
- Calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 3^2 = 9 \] \[ 4ac = 4 \cdot 1 \cdot (-5) = -20 \] \[ b^2 - 4ac = 9 - (-20) = 9 + 20 = 29 \]
- Now substituting back into the quadratic formula:
\[ x = \frac{{-3 \pm \sqrt{29}}}{2 \cdot 1} \] \[ x = \frac{{-3 \pm \sqrt{29}}}{2} \]
- So the solutions are:
\[ x_1 = \frac{{-3 + \sqrt{29}}}{2} \] \[ x_2 = \frac{{-3 - \sqrt{29}}}{2} \]
Thus, the solutions to the equation \( x^2 + 3x - 5 = 0 \) are:
\[ x_1 = \frac{{-3 + \sqrt{29}}}{2}, \quad x_2 = \frac{{-3 - \sqrt{29}}}{2} \]
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