1/3x^2+4=16

Solution for 1/3x^2+4=16 equation:

1/3x^2+4=16

We move all terms to the left:

1/3x^2+4-(16)=0


Domain of the equation: 3x^2!=0

x^2!=0/3

x^2!=√0

x!=0

x∈R


We add all the numbers together, and all the variables

1/3x^2-12=0

We multiply all the terms by the denominator


-12*3x^2+1=0

Wy multiply elements

-36x^2+1=0

a = -36; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-36)·1
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*-36}=\frac{-12}{-72}
=1/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*-36}=\frac{12}{-72}
=-1/6
$