(x+1)(x+2)=(x+3)(x+4)+(x+5)(x+6)

Solution for (x+1)(x+2)=(x+3)(x+4)+(x+5)(x+6) equation:

(x+1)(x+2)=(x+3)(x+4)+(x+5)(x+6)

We move all terms to the left:

(x+1)(x+2)-((x+3)(x+4)+(x+5)(x+6))=0

We multiply parentheses ..

(+x^2+2x+x+2)-((x+3)(x+4)+(x+5)(x+6))=0


We calculate terms in parentheses: -((x+3)(x+4)+(x+5)(x+6)), so:

(x+3)(x+4)+(x+5)(x+6)

We multiply parentheses ..

(+x^2+4x+3x+12)+(x+5)(x+6)

We get rid of parentheses

x^2+4x+3x+(x+5)(x+6)+12

We multiply parentheses ..

x^2+(+x^2+6x+5x+30)+4x+3x+12

We add all the numbers together, and all the variables

x^2+(+x^2+6x+5x+30)+7x+12

We get rid of parentheses

x^2+x^2+6x+5x+7x+30+12

We add all the numbers together, and all the variables

2x^2+18x+42

Back to the equation:

-(2x^2+18x+42)


We get rid of parentheses

x^2-2x^2+2x+x-18x+2-42=0

We add all the numbers together, and all the variables

-1x^2-15x-40=0

a = -1; b = -15; c = -40;
Δ = b2-4ac
Δ = -152-4·(-1)·(-40)
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{65}}{2*-1}=\frac{15-\sqrt{65}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{65}}{2*-1}=\frac{15+\sqrt{65}}{-2}
$