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10p^2-10p=0
a = 10; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·10·0
Δ = 100
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{100}=10$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*10}=\frac{0}{20} =0 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*10}=\frac{20}{20} =1 $
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