(6-v)(4v+1)=0

Solution for (6-v)(4v+1)=0 equation:

(6-v)(4v+1)=0

We add all the numbers together, and all the variables

(-1v+6)(4v+1)=0

We multiply parentheses ..

(-4v^2-1v+24v+6)=0

We get rid of parentheses

-4v^2-1v+24v+6=0

We add all the numbers together, and all the variables

-4v^2+23v+6=0

a = -4; b = 23; c = +6;
Δ = b2-4ac
Δ = 232-4·(-4)·6
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-25}{2*-4}=\frac{-48}{-8}
=+6
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+25}{2*-4}=\frac{2}{-8}
=-1/4
$