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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-9x^2+1=0

a = -9; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-9)·1
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{-(0)-6}{2*-9}=\frac{-6}{-18}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*-9}=\frac{6}{-18}
=-1/3
$