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1/4y+6=1/10y
We move all terms to the left:
1/4y+6-(1/10y)=0
Domain of the equation: 4y!=0
y!=0/4
y!=0
y∈R
Domain of the equation: 10y)!=0We add all the numbers together, and all the variables
y!=0/1
y!=0
y∈R
1/4y-(+1/10y)+6=0
We get rid of parentheses
1/4y-1/10y+6=0
We calculate fractions
10y/40y^2+(-4y)/40y^2+6=0
We multiply all the terms by the denominator
10y+(-4y)+6*40y^2=0
Wy multiply elements
240y^2+10y+(-4y)=0
We get rid of parentheses
240y^2+10y-4y=0
We add all the numbers together, and all the variables
240y^2+6y=0
a = 240; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·240·0
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*240}=\frac{-12}{480} =-1/40 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*240}=\frac{0}{480} =0 $
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