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1/2y+2=1/8y
We move all terms to the left:
1/2y+2-(1/8y)=0
Domain of the equation: 2y!=0
y!=0/2
y!=0
y∈R
Domain of the equation: 8y)!=0We add all the numbers together, and all the variables
y!=0/1
y!=0
y∈R
1/2y-(+1/8y)+2=0
We get rid of parentheses
1/2y-1/8y+2=0
We calculate fractions
8y/16y^2+(-2y)/16y^2+2=0
We multiply all the terms by the denominator
8y+(-2y)+2*16y^2=0
Wy multiply elements
32y^2+8y+(-2y)=0
We get rid of parentheses
32y^2+8y-2y=0
We add all the numbers together, and all the variables
32y^2+6y=0
a = 32; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·32·0
Δ = 36
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*32}=\frac{-12}{64} =-3/16 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*32}=\frac{0}{64} =0 $
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