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

Solution for 2x*2x-((x^2+1)*2)=0 equation:

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

Wy multiply elements

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


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

(x^2+1)*2

We multiply parentheses

2x^2+2

Back to the equation:

-(2x^2+2)


We get rid of parentheses

4x^2-2x^2-2=0

We add all the numbers together, and all the variables

2x^2-2=0

a = 2; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·2·(-2)
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{-(0)-4}{2*2}=\frac{-4}{4}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*2}=\frac{4}{4}
=1
$