2v^2-2v+12=(v+2)(v+2)

Solution for 2v^2-2v+12=(v+2)(v+2) equation:

2v^2-2v+12=(v+2)(v+2)

We move all terms to the left:

2v^2-2v+12-((v+2)(v+2))=0

We multiply parentheses ..

2v^2-((+v^2+2v+2v+4))-2v+12=0


We calculate terms in parentheses: -((+v^2+2v+2v+4)), so:

(+v^2+2v+2v+4)

We get rid of parentheses

v^2+2v+2v+4

We add all the numbers together, and all the variables

v^2+4v+4

Back to the equation:

-(v^2+4v+4)


We add all the numbers together, and all the variables

2v^2-2v-(v^2+4v+4)+12=0

We get rid of parentheses

2v^2-v^2-2v-4v-4+12=0

We add all the numbers together, and all the variables

v^2-6v+8=0

a = 1; b = -6; c = +8;
Δ = b2-4ac
Δ = -62-4·1·8
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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