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

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

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

We move all terms to the left:

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

We multiply parentheses

6x-((x+1)(x+4))-4=0

We multiply parentheses ..

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


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

(+x^2+4x+x+4)

We get rid of parentheses

x^2+4x+x+4

We add all the numbers together, and all the variables

x^2+5x+4

Back to the equation:

-(x^2+5x+4)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2+x-8=0

a = -1; b = 1; c = -8;
Δ = b2-4ac
Δ = 12-4·(-1)·(-8)
Δ = -31
Delta is less than zero, so there is no solution for the equation