(1/3)(6x+27)=12

Solution for (1/3)(6x+27)=12 equation:

(1/3)(6x+27)=12

We move all terms to the left:

(1/3)(6x+27)-(12)=0


Domain of the equation: 3)(6x+27)!=0

x∈R


We add all the numbers together, and all the variables

(+1/3)(6x+27)-12=0

We multiply parentheses ..

(+6x^2+1/3*27)-12=0

We multiply all the terms by the denominator


(+6x^2+1-12*3*27)=0

We get rid of parentheses

6x^2+1-12*3*27=0

We add all the numbers together, and all the variables

6x^2-971=0

a = 6; b = 0; c = -971;
Δ = b2-4ac
Δ = 02-4·6·(-971)
Δ = 23304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{23304}=\sqrt{4*5826}=\sqrt{4}*\sqrt{5826}=2\sqrt{5826}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{5826}}{2*6}=\frac{0-2\sqrt{5826}}{12}
=-\frac{2\sqrt{5826}}{12}
=-\frac{\sqrt{5826}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{5826}}{2*6}=\frac{0+2\sqrt{5826}}{12}
=\frac{2\sqrt{5826}}{12}
=\frac{\sqrt{5826}}{6}
$