8y+3y(1/3y-6)=9y+2y+12

Solution for 8y+3y(1/3y-6)=9y+2y+12 equation:

8y+3y(1/3y-6)=9y+2y+12

We move all terms to the left:

8y+3y(1/3y-6)-(9y+2y+12)=0


Domain of the equation: 3y-6)!=0

y∈R


We add all the numbers together, and all the variables

8y+3y(1/3y-6)-(11y+12)=0

We multiply parentheses

3y^2+8y-18y-(11y+12)=0

We get rid of parentheses

3y^2+8y-18y-11y-12=0

We add all the numbers together, and all the variables

3y^2-21y-12=0

a = 3; b = -21; c = -12;
Δ = b2-4ac
Δ = -212-4·3·(-12)
Δ = 585
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{585}=\sqrt{9*65}=\sqrt{9}*\sqrt{65}=3\sqrt{65}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-3\sqrt{65}}{2*3}=\frac{21-3\sqrt{65}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+3\sqrt{65}}{2*3}=\frac{21+3\sqrt{65}}{6}
$