40-4x=-8x(-7+3x)+4x

Solution for 40-4x=-8x(-7+3x)+4x equation:

40-4x=-8x(-7+3x)+4x

We move all terms to the left:

40-4x-(-8x(-7+3x)+4x)=0

We add all the numbers together, and all the variables

-4x-(-8x(3x-7)+4x)+40=0


We calculate terms in parentheses: -(-8x(3x-7)+4x), so:

-8x(3x-7)+4x

We add all the numbers together, and all the variables

4x-8x(3x-7)

We multiply parentheses

-24x^2+4x+56x

We add all the numbers together, and all the variables

-24x^2+60x

Back to the equation:

-(-24x^2+60x)


We get rid of parentheses

24x^2-60x-4x+40=0

We add all the numbers together, and all the variables

24x^2-64x+40=0

a = 24; b = -64; c = +40;
Δ = b2-4ac
Δ = -642-4·24·40
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{-(-64)-16}{2*24}=\frac{48}{48}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+16}{2*24}=\frac{80}{48}
=1+2/3
$