(6+v)(4v+9)=0

Solution for (6+v)(4v+9)=0 equation:

(6+v)(4v+9)=0

We add all the numbers together, and all the variables

(v+6)(4v+9)=0

We multiply parentheses ..

(+4v^2+9v+24v+54)=0

We get rid of parentheses

4v^2+9v+24v+54=0

We add all the numbers together, and all the variables

4v^2+33v+54=0

a = 4; b = 33; c = +54;
Δ = b2-4ac
Δ = 332-4·4·54
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-15}{2*4}=\frac{-48}{8}
=-6
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+15}{2*4}=\frac{-18}{8}
=-2+1/4
$