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

Solution for (x+1)(4x+5)-(x+2)(x+1)=0 equation:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

4x^2-(+x^2+x+2x+2)+9x+5=0

We get rid of parentheses

4x^2-x^2-x-2x+9x-2+5=0

We add all the numbers together, and all the variables

3x^2+6x+3=0

a = 3; b = 6; c = +3;
Δ = b2-4ac
Δ = 62-4·3·3
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{-6}{6}=-1$