128=(x-4)(x-4)2

Solution for 128=(x-4)(x-4)2 equation:

128=(x-4)(x-4)2

We move all terms to the left:

128-((x-4)(x-4)2)=0

We multiply parentheses ..

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


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

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

We multiply parentheses

2x^2-8x-8x+32

We add all the numbers together, and all the variables

2x^2-16x+32

Back to the equation:

-(2x^2-16x+32)


We get rid of parentheses

-2x^2+16x-32+128=0

We add all the numbers together, and all the variables

-2x^2+16x+96=0

a = -2; b = 16; c = +96;
Δ = b2-4ac
Δ = 162-4·(-2)·96
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-32}{2*-2}=\frac{-48}{-4}
=+12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+32}{2*-2}=\frac{16}{-4}
=-4
$