(4v^2-v-6)+(3v^2-7v+5)=0

Solution for (4v^2-v-6)+(3v^2-7v+5)=0 equation:

(4v^2-v-6)+(3v^2-7v+5)=0

We get rid of parentheses

4v^2+3v^2-v-7v-6+5=0

We add all the numbers together, and all the variables

7v^2-8v-1=0

a = 7; b = -8; c = -1;
Δ = b2-4ac
Δ = -82-4·7·(-1)
Δ = 92
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:

$\sqrt{\Delta}=\sqrt{92}=\sqrt{4*23}=\sqrt{4}*\sqrt{23}=2\sqrt{23}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{23}}{2*7}=\frac{8-2\sqrt{23}}{14}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{23}}{2*7}=\frac{8+2\sqrt{23}}{14}
$