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(3w)2=41
We move all terms to the left:
(3w)2-(41)=0
We add all the numbers together, and all the variables
3w^2-41=0
a = 3; b = 0; c = -41;
Δ = b2-4ac
Δ = 02-4·3·(-41)
Δ = 492
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{492}=\sqrt{4*123}=\sqrt{4}*\sqrt{123}=2\sqrt{123}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{123}}{2*3}=\frac{0-2\sqrt{123}}{6} =-\frac{2\sqrt{123}}{6} =-\frac{\sqrt{123}}{3} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{123}}{2*3}=\frac{0+2\sqrt{123}}{6} =\frac{2\sqrt{123}}{6} =\frac{\sqrt{123}}{3} $
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