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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

(+x^2+x-2x-2)

We get rid of parentheses

x^2+x-2x-2

We add all the numbers together, and all the variables

x^2-1x-2

Back to the equation:

-(x^2-1x-2)


We get rid of parentheses

3x^2-x^2+1x+2=0

We add all the numbers together, and all the variables

2x^2+x+2=0

a = 2; b = 1; c = +2;
Δ = b2-4ac
Δ = 12-4·2·2
Δ = -15
Delta is less than zero, so there is no solution for the equation