(3x-1^2)+2x(1-x)+2=x-7(1-x)x

Solution for (3x-1^2)+2x(1-x)+2=x-7(1-x)x equation:

(3x-1^2)+2x(1-x)+2=x-7(1-x)x

We move all terms to the left:

(3x-1^2)+2x(1-x)+2-(x-7(1-x)x)=0

We add all the numbers together, and all the variables

(3x-1^2)+2x(-1x+1)-(x-7(-1x+1)x)+2=0

We multiply parentheses

-2x^2+(3x-1^2)+2x-(x-7(-1x+1)x)+2=0

We get rid of parentheses

-2x^2+3x+2x-(x-7(-1x+1)x)+2-1^2=0


We calculate terms in parentheses: -(x-7(-1x+1)x), so:

x-7(-1x+1)x

We multiply parentheses

7x^2+x-7x

We add all the numbers together, and all the variables

7x^2-6x

Back to the equation:

-(7x^2-6x)


We add all the numbers together, and all the variables

-2x^2+5x-(7x^2-6x)+1=0

We get rid of parentheses

-2x^2-7x^2+5x+6x+1=0

We add all the numbers together, and all the variables

-9x^2+11x+1=0

a = -9; b = 11; c = +1;
Δ = b2-4ac
Δ = 112-4·(-9)·1
Δ = 157
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{157}}{2*-9}=\frac{-11-\sqrt{157}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{157}}{2*-9}=\frac{-11+\sqrt{157}}{-18}
$