1/2r+6=3-r

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Solution for 1/2r+6=3-r equation:



1/2r+6=3-r
We move all terms to the left:
1/2r+6-(3-r)=0
Domain of the equation: 2r!=0
r!=0/2
r!=0
r∈R
We add all the numbers together, and all the variables
1/2r-(-1r+3)+6=0
We get rid of parentheses
1/2r+1r-3+6=0
We multiply all the terms by the denominator
1r*2r-3*2r+6*2r+1=0
Wy multiply elements
2r^2-6r+12r+1=0
We add all the numbers together, and all the variables
2r^2+6r+1=0
a = 2; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·2·1
Δ = 28
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:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{7}}{2*2}=\frac{-6-2\sqrt{7}}{4} $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{7}}{2*2}=\frac{-6+2\sqrt{7}}{4} $

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