(x^2-9)+(x-3)(2-6x)=0

Solution for (x^2-9)+(x-3)(2-6x)=0 equation:

(x^2-9)+(x-3)(2-6x)=0

We add all the numbers together, and all the variables

(x^2-9)+(x-3)(-6x+2)=0

We get rid of parentheses

x^2+(x-3)(-6x+2)-9=0

We multiply parentheses ..

x^2+(-6x^2+2x+18x-6)-9=0

We get rid of parentheses

x^2-6x^2+2x+18x-6-9=0

We add all the numbers together, and all the variables

-5x^2+20x-15=0

a = -5; b = 20; c = -15;
Δ = b2-4ac
Δ = 202-4·(-5)·(-15)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10}{2*-5}=\frac{-30}{-10}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10}{2*-5}=\frac{-10}{-10}
=1
$