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7p^2+12p-4=0
a = 7; b = 12; c = -4;
Δ = b2-4ac
Δ = 122-4·7·(-4)
Δ = 256
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{256}=16$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-16}{2*7}=\frac{-28}{14} =-2 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+16}{2*7}=\frac{4}{14} =2/7 $
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