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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

4v^2-3v+24v-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
$