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

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

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

We move all terms to the left:

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

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

We add all the numbers together, and all the variables

-3x^2+2x+1=0

a = -3; b = 2; c = +1;
Δ = b2-4ac
Δ = 22-4·(-3)·1
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4}{2*-3}=\frac{-6}{-6}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4}{2*-3}=\frac{2}{-6}
=-1/3
$