(3v+2)(v-4)=v^2+v+5

Solution for (3v+2)(v-4)=v^2+v+5 equation:

(3v+2)(v-4)=v^2+v+5

We move all terms to the left:

(3v+2)(v-4)-(v^2+v+5)=0

We get rid of parentheses

-v^2+(3v+2)(v-4)-v-5=0

We multiply parentheses ..

-v^2+(+3v^2-12v+2v-8)-v-5=0

We add all the numbers together, and all the variables

-1v^2+(+3v^2-12v+2v-8)-1v-5=0

We get rid of parentheses

-1v^2+3v^2-12v+2v-1v-8-5=0

We add all the numbers together, and all the variables

2v^2-11v-13=0

a = 2; b = -11; c = -13;
Δ = b2-4ac
Δ = -112-4·2·(-13)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-15}{2*2}=\frac{-4}{4}
=-1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+15}{2*2}=\frac{26}{4}
=6+1/2
$