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3p^2=10p-7
We move all terms to the left:
3p^2-(10p-7)=0
We get rid of parentheses
3p^2-10p+7=0
a = 3; b = -10; c = +7;
Δ = b2-4ac
Δ = -102-4·3·7
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$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4}{2*3}=\frac{6}{6} =1 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4}{2*3}=\frac{14}{6} =2+1/3 $
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