8x+(5x-6-4x+1)=x-x(x-4)

Solution for 8x+(5x-6-4x+1)=x-x(x-4) equation:

8x+(5x-6-4x+1)=x-x(x-4)

We move all terms to the left:

8x+(5x-6-4x+1)-(x-x(x-4))=0

We add all the numbers together, and all the variables

8x+(x-5)-(x-x(x-4))=0

We get rid of parentheses

8x+x-(x-x(x-4))-5=0


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

x-x(x-4)

We multiply parentheses

-x^2+x+4x

We add all the numbers together, and all the variables

-1x^2+5x

Back to the equation:

-(-1x^2+5x)


We add all the numbers together, and all the variables

-(-1x^2+5x)+9x-5=0

We get rid of parentheses

1x^2-5x+9x-5=0

We add all the numbers together, and all the variables

x^2+4x-5=0

a = 1; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·1·(-5)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6}{2*1}=\frac{-10}{2}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6}{2*1}=\frac{2}{2}
=1
$