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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

(-4x^2+x-5)

We get rid of parentheses

-4x^2+x-5

Back to the equation:

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


We add all the numbers together, and all the variables

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

We get rid of parentheses

6x^2+4x^2-x-6x+5+4=0

We add all the numbers together, and all the variables

10x^2-7x+9=0

a = 10; b = -7; c = +9;
Δ = b2-4ac
Δ = -72-4·10·9
Δ = -311
Delta is less than zero, so there is no solution for the equation