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10p^2-3p-1=0
a = 10; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·10·(-1)
Δ = 49
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{49}=7$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*10}=\frac{-4}{20} =-1/5 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*10}=\frac{10}{20} =1/2 $
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