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

Solution for 6x-x(x-4)+(x^2+4)=4(x+5)-x(x-6) equation:

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

We move all terms to the left:

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

We multiply parentheses

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

We get rid of parentheses

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


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

4(x+5)-x(x-6)

We multiply parentheses

-x^2+4x+6x+20

We add all the numbers together, and all the variables

-1x^2+10x+20

Back to the equation:

-(-1x^2+10x+20)


We add all the numbers together, and all the variables

-(-1x^2+10x+20)+10x+4=0

We get rid of parentheses

1x^2-10x+10x-20+4=0

We add all the numbers together, and all the variables

x^2-16=0

a = 1; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·1·(-16)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*1}=\frac{8}{2}
=4
$