(4v-9)(v+1)=0

Solution for (4v-9)(v+1)=0 equation:

(4v-9)(v+1)=0

We multiply parentheses ..

(+4v^2+4v-9v-9)=0

We get rid of parentheses

4v^2+4v-9v-9=0

We add all the numbers together, and all the variables

4v^2-5v-9=0

a = 4; b = -5; c = -9;
Δ = b2-4ac
Δ = -52-4·4·(-9)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*4}=\frac{-8}{8}
=-1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*4}=\frac{18}{8}
=2+1/4
$