6x+(4+x)=x(x+2)-8(2-x)

Solution for 6x+(4+x)=x(x+2)-8(2-x) equation:

6x+(4+x)=x(x+2)-8(2-x)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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


We calculate terms in parentheses: -(x(x+2)-8(-1x+2)), so:

x(x+2)-8(-1x+2)

We multiply parentheses

x^2+2x+8x-16

We add all the numbers together, and all the variables

x^2+10x-16

Back to the equation:

-(x^2+10x-16)


We add all the numbers together, and all the variables

7x-(x^2+10x-16)+4=0

We get rid of parentheses

-x^2+7x-10x+16+4=0

We add all the numbers together, and all the variables

-1x^2-3x+20=0

a = -1; b = -3; c = +20;
Δ = b2-4ac
Δ = -32-4·(-1)·20
Δ = 89
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{-(-3)-\sqrt{89}}{2*-1}=\frac{3-\sqrt{89}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{89}}{2*-1}=\frac{3+\sqrt{89}}{-2}
$