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28p^2+29p+6=0
a = 28; b = 29; c = +6;
Δ = b2-4ac
Δ = 292-4·28·6
Δ = 169
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{169}=13$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-13}{2*28}=\frac{-42}{56} =-3/4 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+13}{2*28}=\frac{-16}{56} =-2/7 $
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