(1/3)(9k+12)=15

Solution for (1/3)(9k+12)=15 equation:

(1/3)(9k+12)=15

We move all terms to the left:

(1/3)(9k+12)-(15)=0


Domain of the equation: 3)(9k+12)!=0

k∈R


We add all the numbers together, and all the variables

(+1/3)(9k+12)-15=0

We multiply parentheses ..

(+9k^2+1/3*12)-15=0

We multiply all the terms by the denominator


(+9k^2+1-15*3*12)=0

We get rid of parentheses

9k^2+1-15*3*12=0

We add all the numbers together, and all the variables

9k^2-539=0

a = 9; b = 0; c = -539;
Δ = b2-4ac
Δ = 02-4·9·(-539)
Δ = 19404
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{19404}=\sqrt{1764*11}=\sqrt{1764}*\sqrt{11}=42\sqrt{11}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-42\sqrt{11}}{2*9}=\frac{0-42\sqrt{11}}{18}
=-\frac{42\sqrt{11}}{18}
=-\frac{7\sqrt{11}}{3}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+42\sqrt{11}}{2*9}=\frac{0+42\sqrt{11}}{18}
=\frac{42\sqrt{11}}{18}
=\frac{7\sqrt{11}}{3}
$