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

Solution for (1-x)2(3+2)=(9x-4)(1-x) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

(-1x+1)25-((9x-4)(-1x+1))=0

We multiply parentheses

-25x-((9x-4)(-1x+1))+25=0

We multiply parentheses ..

-((-9x^2+9x+4x-4))-25x+25=0


We calculate terms in parentheses: -((-9x^2+9x+4x-4)), so:

(-9x^2+9x+4x-4)

We get rid of parentheses

-9x^2+9x+4x-4

We add all the numbers together, and all the variables

-9x^2+13x-4

Back to the equation:

-(-9x^2+13x-4)


We get rid of parentheses

9x^2-13x-25x+4+25=0

We add all the numbers together, and all the variables

9x^2-38x+29=0

a = 9; b = -38; c = +29;
Δ = b2-4ac
Δ = -382-4·9·29
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-20}{2*9}=\frac{18}{18}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+20}{2*9}=\frac{58}{18}
=3+2/9
$