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2p^2-19p+35=0
a = 2; b = -19; c = +35;
Δ = b2-4ac
Δ = -192-4·2·35
Δ = 81
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{81}=9$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-9}{2*2}=\frac{10}{4} =2+1/2 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+9}{2*2}=\frac{28}{4} =7 $
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