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4+1/3r=1r-8
We move all terms to the left:
4+1/3r-(1r-8)=0
Domain of the equation: 3r!=0We add all the numbers together, and all the variables
r!=0/3
r!=0
r∈R
1/3r-(r-8)+4=0
We get rid of parentheses
1/3r-r+8+4=0
We multiply all the terms by the denominator
-r*3r+8*3r+4*3r+1=0
Wy multiply elements
-3r^2+24r+12r+1=0
We add all the numbers together, and all the variables
-3r^2+36r+1=0
a = -3; b = 36; c = +1;
Δ = b2-4ac
Δ = 362-4·(-3)·1
Δ = 1308
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1308}=\sqrt{4*327}=\sqrt{4}*\sqrt{327}=2\sqrt{327}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-2\sqrt{327}}{2*-3}=\frac{-36-2\sqrt{327}}{-6} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+2\sqrt{327}}{2*-3}=\frac{-36+2\sqrt{327}}{-6} $
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