(1/2)(12y-4)=14-10y

Solution for (1/2)(12y-4)=14-10y equation:

(1/2)(12y-4)=14-10y

We move all terms to the left:

(1/2)(12y-4)-(14-10y)=0


Domain of the equation: 2)(12y-4)!=0

y∈R


We add all the numbers together, and all the variables

(+1/2)(12y-4)-(-10y+14)=0

We get rid of parentheses

(+1/2)(12y-4)+10y-14=0

We multiply parentheses ..

(+12y^2+1/2*-4)+10y-14=0

We multiply all the terms by the denominator


(+12y^2+1+10y*2*-4)-14*2*-4)=0

We add all the numbers together, and all the variables

(+12y^2+1+10y*2*-4)=0

We get rid of parentheses

12y^2+10y*2*+1-4=0

We add all the numbers together, and all the variables

12y^2+10y*2*-3=0

Wy multiply elements

12y^2+20y^2-3=0

We add all the numbers together, and all the variables

32y^2-3=0

a = 32; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·32·(-3)
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{6}}{2*32}=\frac{0-8\sqrt{6}}{64}
=-\frac{8\sqrt{6}}{64}
=-\frac{\sqrt{6}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{6}}{2*32}=\frac{0+8\sqrt{6}}{64}
=\frac{8\sqrt{6}}{64}
=\frac{\sqrt{6}}{8}
$