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3w^2=320
We move all terms to the left:
3w^2-(320)=0
a = 3; b = 0; c = -320;
Δ = b2-4ac
Δ = 02-4·3·(-320)
Δ = 3840
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3840}=\sqrt{256*15}=\sqrt{256}*\sqrt{15}=16\sqrt{15}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{15}}{2*3}=\frac{0-16\sqrt{15}}{6} =-\frac{16\sqrt{15}}{6} =-\frac{8\sqrt{15}}{3} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{15}}{2*3}=\frac{0+16\sqrt{15}}{6} =\frac{16\sqrt{15}}{6} =\frac{8\sqrt{15}}{3} $
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