(3v+2)(5-v)=0

Solution for (3v+2)(5-v)=0 equation:

(3v+2)(5-v)=0

We add all the numbers together, and all the variables

(3v+2)(-1v+5)=0

We multiply parentheses ..

(-3v^2+15v-2v+10)=0

We get rid of parentheses

-3v^2+15v-2v+10=0

We add all the numbers together, and all the variables

-3v^2+13v+10=0

a = -3; b = 13; c = +10;
Δ = b2-4ac
Δ = 132-4·(-3)·10
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-17}{2*-3}=\frac{-30}{-6}
=+5
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*-3}=\frac{4}{-6}
=-2/3
$