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48w^2+54w=0
a = 48; b = 54; c = 0;
Δ = b2-4ac
Δ = 542-4·48·0
Δ = 2916
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{2916}=54$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-54}{2*48}=\frac{-108}{96} =-1+1/8 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+54}{2*48}=\frac{0}{96} =0 $
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