1/3y+y-4=32

Solution for 1/3y+y-4=32 equation:

1/3y+y-4=32

We move all terms to the left:

1/3y+y-4-(32)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R


We add all the numbers together, and all the variables

y+1/3y-36=0

We multiply all the terms by the denominator


y*3y-36*3y+1=0

Wy multiply elements

3y^2-108y+1=0

a = 3; b = -108; c = +1;
Δ = b2-4ac
Δ = -1082-4·3·1
Δ = 11652
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:

$\sqrt{\Delta}=\sqrt{11652}=\sqrt{4*2913}=\sqrt{4}*\sqrt{2913}=2\sqrt{2913}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-108)-2\sqrt{2913}}{2*3}=\frac{108-2\sqrt{2913}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-108)+2\sqrt{2913}}{2*3}=\frac{108+2\sqrt{2913}}{6}
$