-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!=0/2

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 = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{13}}{2*4}=\frac{-6-2\sqrt{13}}{8}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{13}}{2*4}=\frac{-6+2\sqrt{13}}{8}
$