320=5w^2

Solution for 320=5w^2 equation:

320=5w^2

We move all terms to the left:

320-(5w^2)=0

a = -5; b = 0; c = +320;
Δ = b2-4ac
Δ = 02-4·(-5)·320
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80}{2*-5}=\frac{-80}{-10}
=+8
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80}{2*-5}=\frac{80}{-10}
=-8
$