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12-2/5r=2r+1
We move all terms to the left:
12-2/5r-(2r+1)=0
Domain of the equation: 5r!=0We get rid of parentheses
r!=0/5
r!=0
r∈R
-2/5r-2r-1+12=0
We multiply all the terms by the denominator
-2r*5r-1*5r+12*5r-2=0
Wy multiply elements
-10r^2-5r+60r-2=0
We add all the numbers together, and all the variables
-10r^2+55r-2=0
a = -10; b = 55; c = -2;
Δ = b2-4ac
Δ = 552-4·(-10)·(-2)
Δ = 2945
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(55)-\sqrt{2945}}{2*-10}=\frac{-55-\sqrt{2945}}{-20} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(55)+\sqrt{2945}}{2*-10}=\frac{-55+\sqrt{2945}}{-20} $
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