1/8r-(6+2r)=-3

Solution for 1/8r-(6+2r)=-3 equation:

1/8r-(6+2r)=-3

We move all terms to the left:

1/8r-(6+2r)-(-3)=0


Domain of the equation: 8r!=0

r!=0/8

r!=0

r∈R


We add all the numbers together, and all the variables

1/8r-(2r+6)-(-3)=0

We add all the numbers together, and all the variables

1/8r-(2r+6)+3=0

We get rid of parentheses

1/8r-2r-6+3=0

We multiply all the terms by the denominator


-2r*8r-6*8r+3*8r+1=0

Wy multiply elements

-16r^2-48r+24r+1=0

We add all the numbers together, and all the variables

-16r^2-24r+1=0

a = -16; b = -24; c = +1;
Δ = b2-4ac
Δ = -242-4·(-16)·1
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-8\sqrt{10}}{2*-16}=\frac{24-8\sqrt{10}}{-32}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+8\sqrt{10}}{2*-16}=\frac{24+8\sqrt{10}}{-32}
$