Z=1/4y+y+2

Solution for Z=1/4y+y+2 equation:

=1/4Z+Z+2

We move all terms to the left:

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


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

Z∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

-Z-1/4Z-2=0

We multiply all the terms by the denominator


-Z*4Z-2*4Z-1=0

Wy multiply elements

-4Z^2-8Z-1=0

a = -4; b = -8; c = -1;
Δ = b2-4ac
Δ = -82-4·(-4)·(-1)
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{3}}{2*-4}=\frac{8-4\sqrt{3}}{-8}
$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{3}}{2*-4}=\frac{8+4\sqrt{3}}{-8}
$