1/2r+6=3-2-r

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

1/2r+6=3-2-r

We move all terms to the left:

1/2r+6-(3-2-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+1)+6=0

We get rid of parentheses

1/2r+1r-1+6=0

We multiply all the terms by the denominator


1r*2r-1*2r+6*2r+1=0

Wy multiply elements

2r^2-2r+12r+1=0

We add all the numbers together, and all the variables

2r^2+10r+1=0

a = 2; b = 10; c = +1;
Δ = b2-4ac
Δ = 102-4·2·1
Δ = 92
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:

$\sqrt{\Delta}=\sqrt{92}=\sqrt{4*23}=\sqrt{4}*\sqrt{23}=2\sqrt{23}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{23}}{2*2}=\frac{-10-2\sqrt{23}}{4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{23}}{2*2}=\frac{-10+2\sqrt{23}}{4}
$