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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

2x^2-6x-((-1x+1)(x+4))=0

We multiply parentheses ..

2x^2-((-1x^2-4x+x+4))-6x=0


We calculate terms in parentheses: -((-1x^2-4x+x+4)), so:

(-1x^2-4x+x+4)

We get rid of parentheses

-1x^2-4x+x+4

We add all the numbers together, and all the variables

-1x^2-3x+4

Back to the equation:

-(-1x^2-3x+4)


We get rid of parentheses

2x^2+1x^2+3x-6x-4=0

We add all the numbers together, and all the variables

3x^2-3x-4=0

a = 3; b = -3; c = -4;
Δ = b2-4ac
Δ = -32-4·3·(-4)
Δ = 57
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{57}}{2*3}=\frac{3-\sqrt{57}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{57}}{2*3}=\frac{3+\sqrt{57}}{6}
$