(3x+2)(x+1)=(x+4)(x-5)

Solution for (3x+2)(x+1)=(x+4)(x-5) equation:

(3x+2)(x+1)=(x+4)(x-5)

We move all terms to the left:

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

We multiply parentheses ..

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


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

(x+4)(x-5)

We multiply parentheses ..

(+x^2-5x+4x-20)

We get rid of parentheses

x^2-5x+4x-20

We add all the numbers together, and all the variables

x^2-1x-20

Back to the equation:

-(x^2-1x-20)


We get rid of parentheses

3x^2-x^2+3x+2x+1x+2+20=0

We add all the numbers together, and all the variables

2x^2+6x+22=0

a = 2; b = 6; c = +22;
Δ = b2-4ac
Δ = 62-4·2·22
Δ = -140
Delta is less than zero, so there is no solution for the equation