Wx2+(22x2)=120

Solution for Wx2+(22x2)=120 equation:

W2+(22W^2)=120

We move all terms to the left:

W2+(22W^2)-(120)=0

determiningTheFunctionDomain
22W^2+W2-120=0

We add all the numbers together, and all the variables

23W^2-120=0

a = 23; b = 0; c = -120;
Δ = b2-4ac
Δ = 02-4·23·(-120)
Δ = 11040
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:

$\sqrt{\Delta}=\sqrt{11040}=\sqrt{16*690}=\sqrt{16}*\sqrt{690}=4\sqrt{690}$
$W_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{690}}{2*23}=\frac{0-4\sqrt{690}}{46}
=-\frac{4\sqrt{690}}{46}
=-\frac{2\sqrt{690}}{23}
$
$W_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{690}}{2*23}=\frac{0+4\sqrt{690}}{46}
=\frac{4\sqrt{690}}{46}
=\frac{2\sqrt{690}}{23}
$