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1000=150p-5p^2
We move all terms to the left:
1000-(150p-5p^2)=0
We get rid of parentheses
5p^2-150p+1000=0
a = 5; b = -150; c = +1000;
Δ = b2-4ac
Δ = -1502-4·5·1000
Δ = 2500
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{2500}=50$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-150)-50}{2*5}=\frac{100}{10} =10 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-150)+50}{2*5}=\frac{200}{10} =20 $
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