1/3w^2+4=12

Solution for 1/3w^2+4=12 equation:

1/3w^2+4=12

We move all terms to the left:

1/3w^2+4-(12)=0


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

w^2!=0/3

w^2!=√0

w!=0

w∈R


We add all the numbers together, and all the variables

1/3w^2-8=0

We multiply all the terms by the denominator


-8*3w^2+1=0

Wy multiply elements

-24w^2+1=0

a = -24; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-24)·1
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*-24}=\frac{0-4\sqrt{6}}{-48}
=-\frac{4\sqrt{6}}{-48}
=-\frac{\sqrt{6}}{-12}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*-24}=\frac{0+4\sqrt{6}}{-48}
=\frac{4\sqrt{6}}{-48}
=\frac{\sqrt{6}}{-12}
$