(1-3x)(1-3x)=4

Solution for (1-3x)(1-3x)=4 equation:

(1-3x)(1-3x)=4

We move all terms to the left:

(1-3x)(1-3x)-(4)=0

We add all the numbers together, and all the variables

(-3x+1)(-3x+1)-4=0

We multiply parentheses ..

(+9x^2-3x-3x+1)-4=0

We get rid of parentheses

9x^2-3x-3x+1-4=0

We add all the numbers together, and all the variables

9x^2-6x-3=0

a = 9; b = -6; c = -3;
Δ = b2-4ac
Δ = -62-4·9·(-3)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-12}{2*9}=\frac{-6}{18}
=-1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+12}{2*9}=\frac{18}{18}
=1
$