(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 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)-200=0

We get rid of parentheses

9x^2+1x^2+1x-x+3(-2x+1)(2x+1)-1-200=0

We multiply parentheses ..

9x^2+1x^2+3(-4x^2-2x+2x+1)+1x-x-1-200=0

We add all the numbers together, and all the variables

10x^2+3(-4x^2-2x+2x+1)-201=0

We multiply parentheses

10x^2-12x^2-6x+6x+3-201=0

We add all the numbers together, and all the variables

-2x^2-198=0

a = -2; b = 0; c = -198;
Δ = b2-4ac
Δ = 02-4·(-2)·(-198)
Δ = -1584
Delta is less than zero, so there is no solution for the equation