1/3y+10=y-4

Solution for 1/3y+10=y-4 equation:

1/3y+10=y-4

We move all terms to the left:

1/3y+10-(y-4)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R


We get rid of parentheses

1/3y-y+4+10=0

We multiply all the terms by the denominator


-y*3y+4*3y+10*3y+1=0

Wy multiply elements

-3y^2+12y+30y+1=0

We add all the numbers together, and all the variables

-3y^2+42y+1=0

a = -3; b = 42; c = +1;
Δ = b2-4ac
Δ = 422-4·(-3)·1
Δ = 1776
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:

$\sqrt{\Delta}=\sqrt{1776}=\sqrt{16*111}=\sqrt{16}*\sqrt{111}=4\sqrt{111}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-4\sqrt{111}}{2*-3}=\frac{-42-4\sqrt{111}}{-6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+4\sqrt{111}}{2*-3}=\frac{-42+4\sqrt{111}}{-6}
$