(v+9)(5v+3)=0

Solution for (v+9)(5v+3)=0 equation:

(v+9)(5v+3)=0

We multiply parentheses ..

(+5v^2+3v+45v+27)=0

We get rid of parentheses

5v^2+3v+45v+27=0

We add all the numbers together, and all the variables

5v^2+48v+27=0

a = 5; b = 48; c = +27;
Δ = b2-4ac
Δ = 482-4·5·27
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-42}{2*5}=\frac{-90}{10}
=-9
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+42}{2*5}=\frac{-6}{10}
=-3/5
$