(8+v)(5v+2)=0

Solution for (8+v)(5v+2)=0 equation:

(8+v)(5v+2)=0

We add all the numbers together, and all the variables

(v+8)(5v+2)=0

We multiply parentheses ..

(+5v^2+2v+40v+16)=0

We get rid of parentheses

5v^2+2v+40v+16=0

We add all the numbers together, and all the variables

5v^2+42v+16=0

a = 5; b = 42; c = +16;
Δ = b2-4ac
Δ = 422-4·5·16
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-38}{2*5}=\frac{-80}{10}
=-8
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+38}{2*5}=\frac{-4}{10}
=-2/5
$