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

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

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

We move all terms to the left:

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

We multiply parentheses

2x-((x+4)(x+2))+6-1=0

We multiply parentheses ..

-((+x^2+2x+4x+8))+2x+6-1=0


We calculate terms in parentheses: -((+x^2+2x+4x+8)), so:

(+x^2+2x+4x+8)

We get rid of parentheses

x^2+2x+4x+8

We add all the numbers together, and all the variables

x^2+6x+8

Back to the equation:

-(x^2+6x+8)


We add all the numbers together, and all the variables

2x-(x^2+6x+8)+5=0

We get rid of parentheses

-x^2+2x-6x-8+5=0

We add all the numbers together, and all the variables

-1x^2-4x-3=0

a = -1; b = -4; c = -3;
Δ = b2-4ac
Δ = -42-4·(-1)·(-3)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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