For this case we have the following equation:
[tex]2v ^ 2-12 = -12v[/tex]
Rewriting we have:
[tex]2v ^ 2 + 12v-12 = 0[/tex]
Dividing by 2 to both sides of the equation:
[tex]v ^ 2 + 6v-6 = 0[/tex]
We apply the quadratic formula:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
We have to:
[tex]a = 1\\b = 6\\c = -6[/tex]
Substituting:
[tex]x = \frac {-6 \pm \sqrt {6 ^ 2-4 (1) (- 6)}} {2 (1)}\\x = \frac {-6 \pm \sqrt {36 + 24}} {2}\\x = \frac {-6 \pm \sqrt {60}} {2}\\x = \frac {-6 \pm \sqrt {4 * 15}} {2}\\x = \frac {-6 \pm2 \sqrt {15}} {2}[/tex]
Thus, we have two roots:
[tex]x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}[/tex]
ANswer:
[tex]x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}[/tex]
