(3v-7)(v-5)=0

Solution for (3v-7)(v-5)=0 equation:

(3v-7)(v-5)=0

We multiply parentheses ..

(+3v^2-15v-7v+35)=0

We get rid of parentheses

3v^2-15v-7v+35=0

We add all the numbers together, and all the variables

3v^2-22v+35=0

a = 3; b = -22; c = +35;
Δ = b2-4ac
Δ = -222-4·3·35
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-8}{2*3}=\frac{14}{6}
=2+1/3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+8}{2*3}=\frac{30}{6}
=5
$