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

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

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

We multiply parentheses ..

(+3v^2+9v-4v-12)=0

We get rid of parentheses

3v^2+9v-4v-12=0

We add all the numbers together, and all the variables

3v^2+5v-12=0

a = 3; b = 5; c = -12;
Δ = b2-4ac
Δ = 52-4·3·(-12)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*3}=\frac{-18}{6}
=-3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*3}=\frac{8}{6}
=1+1/3
$