For this case we have the following quadratic equation:
[tex]x ^ 2 + 8x = 7\\x ^ 2 + 8x-7 = 0[/tex]
Where:
[tex]a = 1\\b = 8\\c = -7[/tex]
We find the solution by:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
Substituting the values we have:
[tex]x = \frac {-8 \pm \sqrt {8 ^ 2-4 (1) (- 7)}} {2 (1)}\\x = \frac {-8 \pm \sqrt {64 + 28}} {2}\\x = \frac {-8 \pm \sqrt {92}} {2}\\x = \frac {-8 \pm \sqrt {2 ^ 2 * 23}} {2}\\x = \frac {-8 \pm2 \sqrt {23}} {2}\\x = -4 \pm \sqrt {23}[/tex]
Thus, we have two roots:
[tex]x_ {1} = - 4+ \sqrt {23}\\x_ {2} = - 4- \sqrt {23}[/tex]
Answer:
[tex]x_ {1} = - 4+ \sqrt {23}\\x_ {2} = - 4- \sqrt {23}[/tex]
