(5v+1)(2v+9)=0

Solution for (5v+1)(2v+9)=0 equation:

(5v+1)(2v+9)=0

We multiply parentheses ..

(+10v^2+45v+2v+9)=0

We get rid of parentheses

10v^2+45v+2v+9=0

We add all the numbers together, and all the variables

10v^2+47v+9=0

a = 10; b = 47; c = +9;
Δ = b2-4ac
Δ = 472-4·10·9
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(47)-43}{2*10}=\frac{-90}{20}
=-4+1/2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+43}{2*10}=\frac{-4}{20}
=-1/5
$