(v+2)(v-1)=55

Solution for (v+2)(v-1)=55 equation:

(v+2)(v-1)=55

We move all terms to the left:

(v+2)(v-1)-(55)=0

We multiply parentheses ..

(+v^2-1v+2v-2)-55=0

We get rid of parentheses

v^2-1v+2v-2-55=0

We add all the numbers together, and all the variables

v^2+v-57=0

a = 1; b = 1; c = -57;
Δ = b2-4ac
Δ = 12-4·1·(-57)
Δ = 229
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{229}}{2*1}=\frac{-1-\sqrt{229}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{229}}{2*1}=\frac{-1+\sqrt{229}}{2}
$