648^3=(2x-16)(x-8)4

Solution for 648^3=(2x-16)(x-8)4 equation:

648^3=(2x-16)(x-8)4

We move all terms to the left:

648^3-((2x-16)(x-8)4)=0

We add all the numbers together, and all the variables

-((2x-16)(x-8)4)+272097792=0

We multiply parentheses ..

-((+2x^2-16x-16x+128)4)+272097792=0


We calculate terms in parentheses: -((+2x^2-16x-16x+128)4), so:

(+2x^2-16x-16x+128)4

We multiply parentheses

8x^2-64x-64x+512

We add all the numbers together, and all the variables

8x^2-128x+512

Back to the equation:

-(8x^2-128x+512)


We get rid of parentheses

-8x^2+128x-512+272097792=0

We add all the numbers together, and all the variables

-8x^2+128x+272097280=0

a = -8; b = 128; c = +272097280;
Δ = b2-4ac
Δ = 1282-4·(-8)·272097280
Δ = 8707129344
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8707129344}=93312$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-93312}{2*-8}=\frac{-93440}{-16}
=+5840
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+93312}{2*-8}=\frac{93184}{-16}
=-5824
$