4x+2x^2-(3-3x)+x^2+4=6x^2+4-(x^2-5)+4x

Solution for 4x+2x^2-(3-3x)+x^2+4=6x^2+4-(x^2-5)+4x equation:

4x+2x^2-(3-3x)+x^2+4=6x^2+4-(x^2-5)+4x

We move all terms to the left:

4x+2x^2-(3-3x)+x^2+4-(6x^2+4-(x^2-5)+4x)=0

We add all the numbers together, and all the variables

2x^2+x^2+4x-(-3x+3)-(6x^2+4-(x^2-5)+4x)+4=0

We add all the numbers together, and all the variables

3x^2+4x-(-3x+3)-(6x^2+4-(x^2-5)+4x)+4=0

We get rid of parentheses

3x^2+4x+3x-(6x^2+4-(x^2-5)+4x)-3+4=0


We calculate terms in parentheses: -(6x^2+4-(x^2-5)+4x), so:

6x^2+4-(x^2-5)+4x

determiningTheFunctionDomain
6x^2-(x^2-5)+4x+4

We add all the numbers together, and all the variables

6x^2+4x-(x^2-5)+4

We get rid of parentheses

6x^2-x^2+4x+5+4

We add all the numbers together, and all the variables

5x^2+4x+9

Back to the equation:

-(5x^2+4x+9)


We add all the numbers together, and all the variables

3x^2+7x-(5x^2+4x+9)+1=0

We get rid of parentheses

3x^2-5x^2+7x-4x-9+1=0

We add all the numbers together, and all the variables

-2x^2+3x-8=0

a = -2; b = 3; c = -8;
Δ = b2-4ac
Δ = 32-4·(-2)·(-8)
Δ = -55
Delta is less than zero, so there is no solution for the equation