(6v+5)(v+4)=0

Solution for (6v+5)(v+4)=0 equation:

(6v+5)(v+4)=0

We multiply parentheses ..

(+6v^2+24v+5v+20)=0

We get rid of parentheses

6v^2+24v+5v+20=0

We add all the numbers together, and all the variables

6v^2+29v+20=0

a = 6; b = 29; c = +20;
Δ = b2-4ac
Δ = 292-4·6·20
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-19}{2*6}=\frac{-48}{12}
=-4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+19}{2*6}=\frac{-10}{12}
=-5/6
$