(v-6)(4v+1)=0

Solution for (v-6)(4v+1)=0 equation:

(v-6)(4v+1)=0

We multiply parentheses ..

(+4v^2+v-24v-6)=0

We get rid of parentheses

4v^2+v-24v-6=0

We add all the numbers together, and all the variables

4v^2-23v-6=0

a = 4; b = -23; c = -6;
Δ = b2-4ac
Δ = -232-4·4·(-6)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-25}{2*4}=\frac{-2}{8}
=-1/4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+25}{2*4}=\frac{48}{8}
=6
$