4+(w^2)+4(w^2/2)=108

Solution for 4+(w^2)+4(w^2/2)=108 equation:

4+(w^2)+4(w^2/2)=108

We move all terms to the left:

4+(w^2)+4(w^2/2)-(108)=0

We add all the numbers together, and all the variables

w^2+4(w^2/2)-104=0

We multiply parentheses

w^2+4w^2-104=0

We add all the numbers together, and all the variables

5w^2-104=0

a = 5; b = 0; c = -104;
Δ = b2-4ac
Δ = 02-4·5·(-104)
Δ = 2080
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:

$\sqrt{\Delta}=\sqrt{2080}=\sqrt{16*130}=\sqrt{16}*\sqrt{130}=4\sqrt{130}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{130}}{2*5}=\frac{0-4\sqrt{130}}{10}
=-\frac{4\sqrt{130}}{10}
=-\frac{2\sqrt{130}}{5}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{130}}{2*5}=\frac{0+4\sqrt{130}}{10}
=\frac{4\sqrt{130}}{10}
=\frac{2\sqrt{130}}{5}
$