1v-4+2v+3=10v2-v-12

Solution for 1v-4+2v+3=10v2-v-12 equation:

1v-4+2v+3=10v^2-v-12

We move all terms to the left:

1v-4+2v+3-(10v^2-v-12)=0

We add all the numbers together, and all the variables

3v-(10v^2-v-12)-1=0

We get rid of parentheses

-10v^2+3v+v+12-1=0

We add all the numbers together, and all the variables

-10v^2+4v+11=0

a = -10; b = 4; c = +11;
Δ = b2-4ac
Δ = 42-4·(-10)·11
Δ = 456
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:

$\sqrt{\Delta}=\sqrt{456}=\sqrt{4*114}=\sqrt{4}*\sqrt{114}=2\sqrt{114}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{114}}{2*-10}=\frac{-4-2\sqrt{114}}{-20}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{114}}{2*-10}=\frac{-4+2\sqrt{114}}{-20}
$