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28p^2-33p-28=0
a = 28; b = -33; c = -28;
Δ = b2-4ac
Δ = -332-4·28·(-28)
Δ = 4225
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{4225}=65$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-65}{2*28}=\frac{-32}{56} =-4/7 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+65}{2*28}=\frac{98}{56} =1+3/4 $
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