2*(4-x)=7x-3x*(x-2)

Solution for 2*(4-x)=7x-3x*(x-2) equation:

2(4-x)=7x-3x(x-2)

We move all terms to the left:

2(4-x)-(7x-3x(x-2))=0

We add all the numbers together, and all the variables

2(-1x+4)-(7x-3x(x-2))=0

We multiply parentheses

-2x-(7x-3x(x-2))+8=0


We calculate terms in parentheses: -(7x-3x(x-2)), so:

7x-3x(x-2)

We multiply parentheses

-3x^2+7x+6x

We add all the numbers together, and all the variables

-3x^2+13x

Back to the equation:

-(-3x^2+13x)


We get rid of parentheses

3x^2-13x-2x+8=0

We add all the numbers together, and all the variables

3x^2-15x+8=0

a = 3; b = -15; c = +8;
Δ = b2-4ac
Δ = -152-4·3·8
Δ = 129
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{-(-15)-\sqrt{129}}{2*3}=\frac{15-\sqrt{129}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{129}}{2*3}=\frac{15+\sqrt{129}}{6}
$