(3x+1)(4+2x)+12=4-6(x^2+1)

Solution for (3x+1)(4+2x)+12=4-6(x^2+1) equation:

(3x+1)(4+2x)+12=4-6(x^2+1)

We move all terms to the left:

(3x+1)(4+2x)+12-(4-6(x^2+1))=0

We add all the numbers together, and all the variables

(3x+1)(2x+4)-(4-6(x^2+1))+12=0

We multiply parentheses ..

(+6x^2+12x+2x+4)-(4-6(x^2+1))+12=0


We calculate terms in parentheses: -(4-6(x^2+1)), so:

4-6(x^2+1)

determiningTheFunctionDomain
-6(x^2+1)+4

We multiply parentheses

-6x^2-6+4

We add all the numbers together, and all the variables

-6x^2-2

Back to the equation:

-(-6x^2-2)


We get rid of parentheses

6x^2+6x^2+12x+2x+4+2+12=0

We add all the numbers together, and all the variables

12x^2+14x+18=0

a = 12; b = 14; c = +18;
Δ = b2-4ac
Δ = 142-4·12·18
Δ = -668
Delta is less than zero, so there is no solution for the equation