(12/18)+(1/18)r=1

Solution for (12/18)+(1/18)r=1 equation:

(12/18)+(1/18)r=1

We move all terms to the left:

(12/18)+(1/18)r-(1)=0


Domain of the equation: 18)r!=0

r!=0/1

r!=0

r∈R


determiningTheFunctionDomain
(1/18)r-1+(12/18)=0

We add all the numbers together, and all the variables

(+1/18)r-1+(+12/18)=0

We multiply parentheses

r^2-1+(+12/18)=0

We get rid of parentheses

r^2-1+12/18=0

We multiply all the terms by the denominator


r^2*18+12-1*18=0

We add all the numbers together, and all the variables

r^2*18-6=0

Wy multiply elements

18r^2-6=0

a = 18; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·18·(-6)
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{3}}{2*18}=\frac{0-12\sqrt{3}}{36}
=-\frac{12\sqrt{3}}{36}
=-\frac{\sqrt{3}}{3}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{3}}{2*18}=\frac{0+12\sqrt{3}}{36}
=\frac{12\sqrt{3}}{36}
=\frac{\sqrt{3}}{3}
$