(3x-1)(3x+1)-(1-x)(1+x)+3(1-2x)(1+2x)=-199

Solution for (3x-1)(3x+1)-(1-x)(1+x)+3(1-2x)(1+2x)=-199 equation:

(3x-1)(3x+1)-(1-x)(1+x)+3(1-2x)(1+2x)=-199

We move all terms to the left:

(3x-1)(3x+1)-(1-x)(1+x)+3(1-2x)(1+2x)-(-199)=0

We add all the numbers together, and all the variables

(3x-1)(3x+1)-(-1x+1)(x+1)+3(-2x+1)(2x+1)-(-199)=0

We add all the numbers together, and all the variables

(3x-1)(3x+1)-(-1x+1)(x+1)+3(-2x+1)(2x+1)+199=0

We use the square of the difference formula

9x^2-(-1x+1)(x+1)+3(-2x+1)(2x+1)-1+199=0

We multiply parentheses ..

9x^2-(-1x^2-1x+x+1)+3(-2x+1)(2x+1)-1+199=0

We add all the numbers together, and all the variables

9x^2-(-1x^2-1x+x+1)+3(-2x+1)(2x+1)+198=0

We get rid of parentheses

9x^2+1x^2+1x-x+3(-2x+1)(2x+1)-1+198=0

We multiply parentheses ..

9x^2+1x^2+3(-4x^2-2x+2x+1)+1x-x-1+198=0

We add all the numbers together, and all the variables

10x^2+3(-4x^2-2x+2x+1)+197=0

We multiply parentheses

10x^2-12x^2-6x+6x+3+197=0

We add all the numbers together, and all the variables

-2x^2+200=0

a = -2; b = 0; c = +200;
Δ = b2-4ac
Δ = 02-4·(-2)·200
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*-2}=\frac{-40}{-4}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*-2}=\frac{40}{-4}
=-10
$