(4v+3)(6-v)=0

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

(4v+3)(6-v)=0

We add all the numbers together, and all the variables

(4v+3)(-1v+6)=0

We multiply parentheses ..

(-4v^2+24v-3v+18)=0

We get rid of parentheses

-4v^2+24v-3v+18=0

We add all the numbers together, and all the variables

-4v^2+21v+18=0

a = -4; b = 21; c = +18;
Δ = b2-4ac
Δ = 212-4·(-4)·18
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-27}{2*-4}=\frac{-48}{-8}
=+6
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+27}{2*-4}=\frac{6}{-8}
=-3/4
$