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3+3W^2-18W=0
a = 3; b = -18; c = +3;
Δ = b2-4ac
Δ = -182-4·3·3
Δ = 288
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}$$W_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$$W_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-12\sqrt{2}}{2*3}=\frac{18-12\sqrt{2}}{6} $$W_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+12\sqrt{2}}{2*3}=\frac{18+12\sqrt{2}}{6} $
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