(-3v2+6v-7)+(4v2+v+5)=0

Solution for (-3v2+6v-7)+(4v2+v+5)=0 equation:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

v^2+7v-2=0

a = 1; b = 7; c = -2;
Δ = b2-4ac
Δ = 72-4·1·(-2)
Δ = 57
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{57}}{2*1}=\frac{-7-\sqrt{57}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{57}}{2*1}=\frac{-7+\sqrt{57}}{2}
$