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32v^2-41v=0
a = 32; b = -41; c = 0;
Δ = b2-4ac
Δ = -412-4·32·0
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1681}=41$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-41}{2*32}=\frac{0}{64} =0 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+41}{2*32}=\frac{82}{64} =1+9/32 $
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