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10p^2+16p-8=0
a = 10; b = 16; c = -8;
Δ = b2-4ac
Δ = 162-4·10·(-8)
Δ = 576
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{576}=24$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-24}{2*10}=\frac{-40}{20} =-2 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+24}{2*10}=\frac{8}{20} =2/5 $
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