y=1/3y+4=16

Solution for y=1/3y+4=16 equation:

y=1/3y+4=16

We move all terms to the left:

y-(1/3y+4)=0


Domain of the equation: 3y+4)!=0

y∈R


We get rid of parentheses

y-1/3y-4=0

We multiply all the terms by the denominator


y*3y-4*3y-1=0

Wy multiply elements

3y^2-12y-1=0

a = 3; b = -12; c = -1;
Δ = b2-4ac
Δ = -122-4·3·(-1)
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{39}}{2*3}=\frac{12-2\sqrt{39}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{39}}{2*3}=\frac{12+2\sqrt{39}}{6}
$