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p2+43p=0
We add all the numbers together, and all the variables
p^2+43p=0
a = 1; b = 43; c = 0;
Δ = b2-4ac
Δ = 432-4·1·0
Δ = 1849
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{1849}=43$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-43}{2*1}=\frac{-86}{2} =-43 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+43}{2*1}=\frac{0}{2} =0 $
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