(-y+2)=1/2y+4

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

(-y+2)=1/2y+4

We move all terms to the left:

(-y+2)-(1/2y+4)=0


Domain of the equation: 2y+4)!=0

y∈R


We add all the numbers together, and all the variables

(-1y+2)-(1/2y+4)=0

We get rid of parentheses

-1y-1/2y+2-4=0

We multiply all the terms by the denominator


-1y*2y+2*2y-4*2y-1=0

Wy multiply elements

-2y^2+4y-8y-1=0

We add all the numbers together, and all the variables

-2y^2-4y-1=0

a = -2; b = -4; c = -1;
Δ = b2-4ac
Δ = -42-4·(-2)·(-1)
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{2}}{2*-2}=\frac{4-2\sqrt{2}}{-4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{2}}{2*-2}=\frac{4+2\sqrt{2}}{-4}
$