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

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

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

We move all terms to the left:

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


Domain of the equation: 2r)!=0

r!=0/1

r!=0

r∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


2r*2r-3*2r+6*2r+1=0

Wy multiply elements

4r^2-6r+12r+1=0

We add all the numbers together, and all the variables

4r^2+6r+1=0

a = 4; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·4·1
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{5}}{2*4}=\frac{-6-2\sqrt{5}}{8}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{5}}{2*4}=\frac{-6+2\sqrt{5}}{8}
$