(x-8)(x-5)=x-1

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

(x-8)(x-5)=x-1

We move all terms to the left:

(x-8)(x-5)-(x-1)=0

We get rid of parentheses

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

We multiply parentheses ..

(+x^2-5x-8x+40)-x+1=0

We add all the numbers together, and all the variables

(+x^2-5x-8x+40)-1x+1=0

We get rid of parentheses

x^2-5x-8x-1x+40+1=0

We add all the numbers together, and all the variables

x^2-14x+41=0

a = 1; b = -14; c = +41;
Δ = b2-4ac
Δ = -142-4·1·41
Δ = 32
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-4\sqrt{2}}{2*1}=\frac{14-4\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4\sqrt{2}}{2*1}=\frac{14+4\sqrt{2}}{2}
$