x^2+(4x-18)(4x-18)-4x-8(4x-18)=0

Solution for x^2+(4x-18)(4x-18)-4x-8(4x-18)=0 equation:

x^2+(4x-18)(4x-18)-4x-8(4x-18)=0

We add all the numbers together, and all the variables

x^2-4x+(4x-18)(4x-18)-8(4x-18)=0

We multiply parentheses

x^2-4x+(4x-18)(4x-18)-32x+144=0

We multiply parentheses ..

x^2+(+16x^2-72x-72x+324)-4x-32x+144=0

We add all the numbers together, and all the variables

x^2+(+16x^2-72x-72x+324)-36x+144=0

We get rid of parentheses

x^2+16x^2-72x-72x-36x+324+144=0

We add all the numbers together, and all the variables

17x^2-180x+468=0

a = 17; b = -180; c = +468;
Δ = b2-4ac
Δ = -1802-4·17·468
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-180)-24}{2*17}=\frac{156}{34}
=4+10/17
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-180)+24}{2*17}=\frac{204}{34}
=6
$