For this case we have the following quadratic equation:
[tex]-2x ^ 2 -6x + 5 = 0[/tex]
Where:
[tex]a = -2\\b = -6\\c = 5[/tex]
The roots will be given by the following formula:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
Substituting the values we have:
[tex]x = \frac {- (- 6) \pm \sqrt {(- 6) ^ 2-4 (-2) (5)}} {2 (-2)}\\x = \frac {6 \pm \sqrt {36 + 40}} {- 4}\\x = \frac {6 \pm \sqrt {76}} {- 4}\\x = \frac {6 \pm \sqrt {2 ^ 2 * 19}} {- 4}\\x = \frac {6 \pm2 \sqrt {19}} {- 4}\\x = \frac {3 \pm \sqrt {19}} {- 2}[/tex]
We have two roots:
[tex]x_ {1} = \frac {-3- \sqrt {19}} {-2}\\x_ {2} = \frac {-3+ \sqrt {19}} {2}[/tex]
Answer:
[tex]x_ {1} = \frac {-3- \sqrt {19}} {2}\\x_ {2} = \frac {-3+ \sqrt {19}} {2}[/tex]
