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2p^2+4p-30=0
a = 2; b = 4; c = -30;
Δ = b2-4ac
Δ = 42-4·2·(-30)
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$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-16}{2*2}=\frac{-20}{4} =-5 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16}{2*2}=\frac{12}{4} =3 $
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