(n2-3n+12)=(n+4)

Solution for (n2-3n+12)=(n+4) equation:

(n2-3n+12)=(n+4)

We move all terms to the left:

(n2-3n+12)-((n+4))=0

We add all the numbers together, and all the variables

(+n^2-3n+12)-((n+4))=0

We get rid of parentheses

n^2-3n-((n+4))+12=0


We calculate terms in parentheses: -((n+4)), so:

(n+4)

We get rid of parentheses

n+4

Back to the equation:

-(n+4)


We get rid of parentheses

n^2-3n-n-4+12=0

We add all the numbers together, and all the variables

n^2-4n+8=0

a = 1; b = -4; c = +8;
Δ = b2-4ac
Δ = -42-4·1·8
Δ = -16
Delta is less than zero, so there is no solution for the equation