For this case we have the following equation:
![2v ^ 2-12 = -12v](https://tex.z-dn.net/?f=2v%20%5E%202-12%20%3D%20-12v)
Rewriting we have:
![2v ^ 2 + 12v-12 = 0](https://tex.z-dn.net/?f=2v%20%5E%202%20%2B%2012v-12%20%3D%200)
Dividing by 2 to both sides of the equation:
![v ^ 2 + 6v-6 = 0](https://tex.z-dn.net/?f=v%20%5E%202%20%2B%206v-6%20%3D%200)
We apply the quadratic formula:
![x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%20%7B-b%20%5Cpm%20%5Csqrt%20%7Bb%20%5E%202-4%20%28a%29%20%28c%29%7D%7D%20%7B2a%7D)
We have to:
![a = 1\\b = 6\\c = -6](https://tex.z-dn.net/?f=a%20%3D%201%5C%5Cb%20%3D%206%5C%5Cc%20%3D%20-6)
Substituting:
![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}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%20%7B-6%20%5Cpm%20%5Csqrt%20%7B6%20%5E%202-4%20%281%29%20%28-%206%29%7D%7D%20%7B2%20%281%29%7D%5C%5Cx%20%3D%20%5Cfrac%20%7B-6%20%5Cpm%20%5Csqrt%20%7B36%20%2B%2024%7D%7D%20%7B2%7D%5C%5Cx%20%3D%20%5Cfrac%20%7B-6%20%5Cpm%20%5Csqrt%20%7B60%7D%7D%20%7B2%7D%5C%5Cx%20%3D%20%5Cfrac%20%7B-6%20%5Cpm%20%5Csqrt%20%7B4%20%2A%2015%7D%7D%20%7B2%7D%5C%5Cx%20%3D%20%5Cfrac%20%7B-6%20%5Cpm2%20%5Csqrt%20%7B15%7D%7D%20%7B2%7D)
Thus, we have two roots:
![x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}](https://tex.z-dn.net/?f=x_%20%7B1%7D%20%3D%20-%203%2B%20%5Csqrt%20%7B15%7D%5C%5Cx_%20%7B2%7D%20%3D%20-%203-%20%5Csqrt%20%7B15%7D)
ANswer:
![x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}](https://tex.z-dn.net/?f=x_%20%7B1%7D%20%3D%20-%203%2B%20%5Csqrt%20%7B15%7D%5C%5Cx_%20%7B2%7D%20%3D%20-%203-%20%5Csqrt%20%7B15%7D)
It equals 5/6. If you want to know why, just ask me.
FOIL
x^2+11x-9x-99
x^2+2x-99
I am thinking that c is correct and i am truely sorry if u get it wrong