(3v+1)(4v+7)=0

Solution for (3v+1)(4v+7)=0 equation:

(3v+1)(4v+7)=0

We multiply parentheses ..

(+12v^2+21v+4v+7)=0

We get rid of parentheses

12v^2+21v+4v+7=0

We add all the numbers together, and all the variables

12v^2+25v+7=0

a = 12; b = 25; c = +7;
Δ = b2-4ac
Δ = 252-4·12·7
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-17}{2*12}=\frac{-42}{24}
=-1+3/4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+17}{2*12}=\frac{-8}{24}
=-1/3
$