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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+x^2+4x-1x-4)), so:

(+x^2+4x-1x-4)

We get rid of parentheses

x^2+4x-1x-4

We add all the numbers together, and all the variables

x^2+3x-4

Back to the equation:

-(x^2+3x-4)


We add all the numbers together, and all the variables

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

We get rid of parentheses

-x^2+2x-3x+4+2=0

We add all the numbers together, and all the variables

-1x^2-1x+6=0

a = -1; b = -1; c = +6;
Δ = b2-4ac
Δ = -12-4·(-1)·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*-1}=\frac{-4}{-2}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*-1}=\frac{6}{-2}
=-3
$