(x^2-6)-(1-x)(-6)=x

Solution for (x^2-6)-(1-x)(-6)=x equation:

(x^2-6)-(1-x)(-6)=x

We move all terms to the left:

(x^2-6)-(1-x)(-6)-(x)=0

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

x^2-1x-(6x-6)-6=0

We get rid of parentheses

x^2-1x-6x+6-6=0

We add all the numbers together, and all the variables

x^2-7x=0

a = 1; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·1·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*1}=\frac{14}{2}
=7
$