(x-1)-(x+3)(x-3)=8

Solution for (x-1)-(x+3)(x-3)=8 equation:

(x-1)-(x+3)(x-3)=8

We move all terms to the left:

(x-1)-(x+3)(x-3)-(8)=0

We use the square of the difference formula

x^2+(x-1)+9-8=0

We get rid of parentheses

x^2+x-1+9-8=0

We add all the numbers together, and all the variables

x^2+x=0

a = 1; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·1·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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