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

Solution for (4+v)(3v-5)=0 equation:

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

We add all the numbers together, and all the variables

(v+4)(3v-5)=0

We multiply parentheses ..

(+3v^2-5v+12v-20)=0

We get rid of parentheses

3v^2-5v+12v-20=0

We add all the numbers together, and all the variables

3v^2+7v-20=0

a = 3; b = 7; c = -20;
Δ = b2-4ac
Δ = 72-4·3·(-20)
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{-(7)-17}{2*3}=\frac{-24}{6}
=-4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+17}{2*3}=\frac{10}{6}
=1+2/3
$