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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: +(3x^2+4x-1+(-2x^2-3x+2)), so:

3x^2+4x-1+(-2x^2-3x+2)

determiningTheFunctionDomain
3x^2+(-2x^2-3x+2)+4x-1

We get rid of parentheses

3x^2-2x^2-3x+4x+2-1

We add all the numbers together, and all the variables

x^2+x+1

Back to the equation:

+(x^2+x+1)


We get rid of parentheses

x^2+x+1=0

a = 1; b = 1; c = +1;
Δ = b2-4ac
Δ = 12-4·1·1
Δ = -3
Delta is less than zero, so there is no solution for the equation