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2p^2-11p=0
a = 2; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·2·0
Δ = 121
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{121}=11$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*2}=\frac{0}{4} =0 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*2}=\frac{22}{4} =5+1/2 $
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