(x-4)(x-4)+7=128

Solution for (x-4)(x-4)+7=128 equation:

(x-4)(x-4)+7=128

We move all terms to the left:

(x-4)(x-4)+7-(128)=0

We add all the numbers together, and all the variables

(x-4)(x-4)-121=0

We multiply parentheses ..

(+x^2-4x-4x+16)-121=0

We get rid of parentheses

x^2-4x-4x+16-121=0

We add all the numbers together, and all the variables

x^2-8x-105=0

a = 1; b = -8; c = -105;
Δ = b2-4ac
Δ = -82-4·1·(-105)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-22}{2*1}=\frac{-14}{2}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+22}{2*1}=\frac{30}{2}
=15
$