(2v-5)(5v+3)=0

Solution for (2v-5)(5v+3)=0 equation:

(2v-5)(5v+3)=0

We multiply parentheses ..

(+10v^2+6v-25v-15)=0

We get rid of parentheses

10v^2+6v-25v-15=0

We add all the numbers together, and all the variables

10v^2-19v-15=0

a = 10; b = -19; c = -15;
Δ = b2-4ac
Δ = -192-4·10·(-15)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-31}{2*10}=\frac{-12}{20}
=-3/5
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+31}{2*10}=\frac{50}{20}
=2+1/2
$