(-6v^2+2v-1)+(3v^2-9v+6)-(4v^2-9v-6)=0

Solution for (-6v^2+2v-1)+(3v^2-9v+6)-(4v^2-9v-6)=0 equation:

(-6v^2+2v-1)+(3v^2-9v+6)-(4v^2-9v-6)=0

We get rid of parentheses

-6v^2+3v^2-4v^2+2v-9v+9v-1+6+6=0

We add all the numbers together, and all the variables

-7v^2+2v+11=0

a = -7; b = 2; c = +11;
Δ = b2-4ac
Δ = 22-4·(-7)·11
Δ = 312
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{312}=\sqrt{4*78}=\sqrt{4}*\sqrt{78}=2\sqrt{78}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{78}}{2*-7}=\frac{-2-2\sqrt{78}}{-14}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{78}}{2*-7}=\frac{-2+2\sqrt{78}}{-14}
$