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4+2p=10(3/5p-5)
We move all terms to the left:
4+2p-(10(3/5p-5))=0
Domain of the equation: 5p-5))!=0We multiply all the terms by the denominator
p∈R
2p*5p+4*5p-5))-(10(3-5))=0
We add all the numbers together, and all the variables
2p*5p+4*5p-5))-(10(-2))=0
We add all the numbers together, and all the variables
2p*5p+4*5p=0
Wy multiply elements
10p^2+20p=0
a = 10; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·10·0
Δ = 400
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{400}=20$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*10}=\frac{-40}{20} =-2 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*10}=\frac{0}{20} =0 $
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