(5v+4)(7v+4)=0

Solution for (5v+4)(7v+4)=0 equation:

(5v+4)(7v+4)=0

We multiply parentheses ..

(+35v^2+20v+28v+16)=0

We get rid of parentheses

35v^2+20v+28v+16=0

We add all the numbers together, and all the variables

35v^2+48v+16=0

a = 35; b = 48; c = +16;
Δ = b2-4ac
Δ = 482-4·35·16
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{-(48)-8}{2*35}=\frac{-56}{70}
=-4/5
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+8}{2*35}=\frac{-40}{70}
=-4/7
$