For this case we have the following quadratic equation:
[tex]x ^ 2 + 3x-5 = 0[/tex]
Where:
[tex]a = 1\\b = 3\\c = -5[/tex]
We look for the solution using the following formula:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
Substituting the values:
[tex]x = \frac {-3 \pm \sqrt {3 ^ 2-4 (1) (- 5)}} {2 (1)}\\x = \frac {-3 \pm \sqrt {9 + 20}} {2}\\x = \frac {-3 \pm \sqrt {29}} {2}[/tex]
Thus, we have two roots:
[tex]x_ {1} = \frac {-3+ \sqrt {29}} {2}\\x_ {2} = \frac {-3- \sqrt {29}} {2}[/tex]
Answer:
[tex]x_ {1} = \frac {-3+ \sqrt {29}} {2}\\x_ {2} = \frac {-3- \sqrt {29}} {2}[/tex]
