x^2-4x+4=(4x+9)(x-2)

Solution for x^2-4x+4=(4x+9)(x-2) equation:

x^2-4x+4=(4x+9)(x-2)

We move all terms to the left:

x^2-4x+4-((4x+9)(x-2))=0

We multiply parentheses ..

x^2-((+4x^2-8x+9x-18))-4x+4=0


We calculate terms in parentheses: -((+4x^2-8x+9x-18)), so:

(+4x^2-8x+9x-18)

We get rid of parentheses

4x^2-8x+9x-18

We add all the numbers together, and all the variables

4x^2+x-18

Back to the equation:

-(4x^2+x-18)


We add all the numbers together, and all the variables

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

We get rid of parentheses

x^2-4x^2-4x-x+18+4=0

We add all the numbers together, and all the variables

-3x^2-5x+22=0

a = -3; b = -5; c = +22;
Δ = b2-4ac
Δ = -52-4·(-3)·22
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-17}{2*-3}=\frac{-12}{-6}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+17}{2*-3}=\frac{22}{-6}
=-3+2/3
$