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

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

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

We move all terms to the left:

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

We get rid of parentheses

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


We calculate terms in parentheses: -(2x+(4x^2-1)), so:

2x+(4x^2-1)

We get rid of parentheses

4x^2+2x-1

Back to the equation:

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


We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -3; b = -1; c = +2;
Δ = b2-4ac
Δ = -12-4·(-3)·2
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*-3}=\frac{-4}{-6}
=2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*-3}=\frac{6}{-6}
=-1
$