1/2r+2(3/4r-1)=r+6

Solution for 1/2r+2(3/4r-1)=r+6 equation:

1/2r+2(3/4r-1)=r+6

We move all terms to the left:

1/2r+2(3/4r-1)-(r+6)=0


Domain of the equation: 2r!=0

r!=0/2

r!=0

r∈R



Domain of the equation: 4r-1)!=0

r∈R


We multiply parentheses

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

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

We add all the numbers together, and all the variables

10r^2-16r+1=0

a = 10; b = -16; c = +1;
Δ = b2-4ac
Δ = -162-4·10·1
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-6\sqrt{6}}{2*10}=\frac{16-6\sqrt{6}}{20}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+6\sqrt{6}}{2*10}=\frac{16+6\sqrt{6}}{20}
$