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64p^2+40p=0
a = 64; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·64·0
Δ = 1600
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{1600}=40$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40}{2*64}=\frac{-80}{128} =-5/8 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*64}=\frac{0}{128} =0 $
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