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41p^2-19p=0
a = 41; b = -19; c = 0;
Δ = b2-4ac
Δ = -192-4·41·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{361}=19$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-19}{2*41}=\frac{0}{82} =0 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+19}{2*41}=\frac{38}{82} =19/41 $
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