x^2+4x+4=100x^2+20x+1

Solution for x^2+4x+4=100x^2+20x+1 equation:

x^2+4x+4=100x^2+20x+1

We move all terms to the left:

x^2+4x+4-(100x^2+20x+1)=0

We get rid of parentheses

x^2-100x^2+4x-20x-1+4=0

We add all the numbers together, and all the variables

-99x^2-16x+3=0

a = -99; b = -16; c = +3;
Δ = b2-4ac
Δ = -162-4·(-99)·3
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-38}{2*-99}=\frac{-22}{-198}
=1/9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+38}{2*-99}=\frac{54}{-198}
=-3/11
$