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(p-5)+2(5-p)p=2
We move all terms to the left:
(p-5)+2(5-p)p-(2)=0
We add all the numbers together, and all the variables
(p-5)+2(-1p+5)p-2=0
We multiply parentheses
-2p^2+(p-5)+10p-2=0
We get rid of parentheses
-2p^2+p+10p-5-2=0
We add all the numbers together, and all the variables
-2p^2+11p-7=0
a = -2; b = 11; c = -7;
Δ = b2-4ac
Δ = 112-4·(-2)·(-7)
Δ = 65
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{65}}{2*-2}=\frac{-11-\sqrt{65}}{-4} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{65}}{2*-2}=\frac{-11+\sqrt{65}}{-4} $
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