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2r+5=31/2r+2
We move all terms to the left:
2r+5-(31/2r+2)=0
Domain of the equation: 2r+2)!=0We get rid of parentheses
r∈R
2r-31/2r-2+5=0
We multiply all the terms by the denominator
2r*2r-2*2r+5*2r-31=0
Wy multiply elements
4r^2-4r+10r-31=0
We add all the numbers together, and all the variables
4r^2+6r-31=0
a = 4; b = 6; c = -31;
Δ = b2-4ac
Δ = 62-4·4·(-31)
Δ = 532
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{532}=\sqrt{4*133}=\sqrt{4}*\sqrt{133}=2\sqrt{133}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{133}}{2*4}=\frac{-6-2\sqrt{133}}{8} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{133}}{2*4}=\frac{-6+2\sqrt{133}}{8} $
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