8x(4-2x)=4x(3-5x)+4x

Solution for 8x(4-2x)=4x(3-5x)+4x equation:

8x(4-2x)=4x(3-5x)+4x

We move all terms to the left:

8x(4-2x)-(4x(3-5x)+4x)=0

We add all the numbers together, and all the variables

8x(-2x+4)-(4x(-5x+3)+4x)=0

We multiply parentheses

-16x^2+32x-(4x(-5x+3)+4x)=0


We calculate terms in parentheses: -(4x(-5x+3)+4x), so:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

-20x^2+4x+12x

We add all the numbers together, and all the variables

-20x^2+16x

Back to the equation:

-(-20x^2+16x)


We get rid of parentheses

-16x^2+20x^2-16x+32x=0

We add all the numbers together, and all the variables

4x^2+16x=0

a = 4; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·4·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*4}=\frac{-32}{8}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*4}=\frac{0}{8}
=0
$