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=-P^2+130P-3000
We move all terms to the left:
-(-P^2+130P-3000)=0
We get rid of parentheses
P^2-130P+3000=0
a = 1; b = -130; c = +3000;
Δ = b2-4ac
Δ = -1302-4·1·3000
Δ = 4900
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{4900}=70$$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-130)-70}{2*1}=\frac{60}{2} =30 $$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-130)+70}{2*1}=\frac{200}{2} =100 $
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