(2/3)y-(1/6)y+4=19

Solution for (2/3)y-(1/6)y+4=19 equation:

(2/3)y-(1/6)y+4=19

We move all terms to the left:

(2/3)y-(1/6)y+4-(19)=0


Domain of the equation: 3)y!=0

y!=0/1

y!=0

y∈R



Domain of the equation: 6)y!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

(+2/3)y-(+1/6)y+4-19=0

We add all the numbers together, and all the variables

(+2/3)y-(+1/6)y-15=0

We multiply parentheses

2y^2-y^2-15=0

We add all the numbers together, and all the variables

y^2-15=0

a = 1; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·1·(-15)
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{15}}{2*1}=\frac{0-2\sqrt{15}}{2}
=-\frac{2\sqrt{15}}{2}
=-\sqrt{15}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{15}}{2*1}=\frac{0+2\sqrt{15}}{2}
=\frac{2\sqrt{15}}{2}
=\sqrt{15}
$