x^2-4x+4=9x^2+6x+1

Solution for x^2-4x+4=9x^2+6x+1 equation:

x^2-4x+4=9x^2+6x+1

We move all terms to the left:

x^2-4x+4-(9x^2+6x+1)=0

We get rid of parentheses

x^2-9x^2-4x-6x-1+4=0

We add all the numbers together, and all the variables

-8x^2-10x+3=0

a = -8; b = -10; c = +3;
Δ = b2-4ac
Δ = -102-4·(-8)·3
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-14}{2*-8}=\frac{-4}{-16}
=1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+14}{2*-8}=\frac{24}{-16}
=-1+1/2
$