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21w^2-35w=0
a = 21; b = -35; c = 0;
Δ = b2-4ac
Δ = -352-4·21·0
Δ = 1225
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{1225}=35$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-35}{2*21}=\frac{0}{42} =0 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+35}{2*21}=\frac{70}{42} =1+2/3 $
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