(x2+4)2+32=12x2=48

Solution for (x2+4)2+32=12x2=48 equation:

(x2+4)2+32=12x^2=48

We move all terms to the left:

(x2+4)2+32-(12x^2)=0

determiningTheFunctionDomain
-12x^2+(x2+4)2+32=0

We add all the numbers together, and all the variables

-12x^2+(+x^2+4)2+32=0

We multiply parentheses

-12x^2+2x^2+8+32=0

We add all the numbers together, and all the variables

-10x^2+40=0

a = -10; b = 0; c = +40;
Δ = b2-4ac
Δ = 02-4·(-10)·40
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*-10}=\frac{-40}{-20}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*-10}=\frac{40}{-20}
=-2
$