(x2+3)2=4x2=12

Solution for (x2+3)2=4x2=12 equation:

(x2+3)2=4x^2=12

We move all terms to the left:

(x2+3)2-(4x^2)=0

determiningTheFunctionDomain
-4x^2+(x2+3)2=0

We add all the numbers together, and all the variables

-4x^2+(+x^2+3)2=0

We multiply parentheses

-4x^2+2x^2+6=0

We add all the numbers together, and all the variables

-2x^2+6=0

a = -2; b = 0; c = +6;
Δ = b2-4ac
Δ = 02-4·(-2)·6
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*-2}=\frac{0-4\sqrt{3}}{-4}
=-\frac{4\sqrt{3}}{-4}
=-\frac{\sqrt{3}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*-2}=\frac{0+4\sqrt{3}}{-4}
=\frac{4\sqrt{3}}{-4}
=\frac{\sqrt{3}}{-1}
$