(3x-1)(x-3)-9+x^2=0

Solution for (3x-1)(x-3)-9+x^2=0 equation:

(3x-1)(x-3)-9+x^2=0

We multiply parentheses ..

x^2+(+3x^2-9x-1x+3)-9=0

We get rid of parentheses

x^2+3x^2-9x-1x+3-9=0

We add all the numbers together, and all the variables

4x^2-10x-6=0

a = 4; b = -10; c = -6;
Δ = b2-4ac
Δ = -102-4·4·(-6)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-14}{2*4}=\frac{-4}{8}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+14}{2*4}=\frac{24}{8}
=3
$