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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -3; b = 2; c = +5;
Δ = b2-4ac
Δ = 22-4·(-3)·5
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*-3}=\frac{-10}{-6}
=1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*-3}=\frac{6}{-6}
=-1
$