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7p^2-10p+3=0
a = 7; b = -10; c = +3;
Δ = b2-4ac
Δ = -102-4·7·3
Δ = 16
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{16}=4$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4}{2*7}=\frac{6}{14} =3/7 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4}{2*7}=\frac{14}{14} =1 $
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