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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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


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

-x(4x+2)

We multiply parentheses

-4x^2-2x

Back to the equation:

-(-4x^2-2x)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-2x^2+4x=0

a = -2; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-2)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-2}=\frac{-8}{-4}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-2}=\frac{0}{-4}
=0
$