5(1-x)+(x-3)4=(x-1)(1-x)

Solution for 5(1-x)+(x-3)4=(x-1)(1-x) equation:

5(1-x)+(x-3)4=(x-1)(1-x)

We move all terms to the left:

5(1-x)+(x-3)4-((x-1)(1-x))=0

We add all the numbers together, and all the variables

5(-1x+1)+(x-3)4-((x-1)(-1x+1))=0

We multiply parentheses

-5x+4x-((x-1)(-1x+1))+5-12=0

We multiply parentheses ..

-((-1x^2+x+x-1))-5x+4x+5-12=0


We calculate terms in parentheses: -((-1x^2+x+x-1)), so:

(-1x^2+x+x-1)

We get rid of parentheses

-1x^2+x+x-1

We add all the numbers together, and all the variables

-1x^2+2x-1

Back to the equation:

-(-1x^2+2x-1)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-3x-6=0

a = 1; b = -3; c = -6;
Δ = b2-4ac
Δ = -32-4·1·(-6)
Δ = 33
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{-(-3)-\sqrt{33}}{2*1}=\frac{3-\sqrt{33}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{33}}{2*1}=\frac{3+\sqrt{33}}{2}
$