(x-2)*(x+4)=(X+3)+(X+6)+(X+4)

Solution for (x-2)*(x+4)=(X+3)+(X+6)+(X+4) equation:

(x-2)(x+4)=(x+3)+(x+6)+(x+4)

We move all terms to the left:

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

We multiply parentheses ..

(+x^2+4x-2x-8)-((x+3)+(x+6)+(x+4))=0


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

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

We get rid of parentheses

x+x+x+3+6+4

We add all the numbers together, and all the variables

3x+13

Back to the equation:

-(3x+13)


We get rid of parentheses

x^2+4x-2x-3x-8-13=0

We add all the numbers together, and all the variables

x^2-1x-21=0

a = 1; b = -1; c = -21;
Δ = b2-4ac
Δ = -12-4·1·(-21)
Δ = 85
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{-(-1)-\sqrt{85}}{2*1}=\frac{1-\sqrt{85}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{85}}{2*1}=\frac{1+\sqrt{85}}{2}
$