19-(20-5x)=3x(8-4x)

Solution for 19-(20-5x)=3x(8-4x) equation:

19-(20-5x)=3x(8-4x)

We move all terms to the left:

19-(20-5x)-(3x(8-4x))=0

We add all the numbers together, and all the variables

-(-5x+20)-(3x(-4x+8))+19=0

We get rid of parentheses

5x-(3x(-4x+8))-20+19=0


We calculate terms in parentheses: -(3x(-4x+8)), so:

3x(-4x+8)

We multiply parentheses

-12x^2+24x

Back to the equation:

-(-12x^2+24x)


We add all the numbers together, and all the variables

-(-12x^2+24x)+5x-1=0

We get rid of parentheses

12x^2-24x+5x-1=0

We add all the numbers together, and all the variables

12x^2-19x-1=0

a = 12; b = -19; c = -1;
Δ = b2-4ac
Δ = -192-4·12·(-1)
Δ = 409
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{-(-19)-\sqrt{409}}{2*12}=\frac{19-\sqrt{409}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{409}}{2*12}=\frac{19+\sqrt{409}}{24}
$