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1/2y+5-4/4y=0
Domain of the equation: 2y!=0
y!=0/2
y!=0
y∈R
Domain of the equation: 4y!=0We calculate fractions
y!=0/4
y!=0
y∈R
4y/8y^2+(-8y)/8y^2+5=0
We multiply all the terms by the denominator
4y+(-8y)+5*8y^2=0
Wy multiply elements
40y^2+4y+(-8y)=0
We get rid of parentheses
40y^2+4y-8y=0
We add all the numbers together, and all the variables
40y^2-4y=0
a = 40; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·40·0
Δ = 16
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{16}=4$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*40}=\frac{0}{80} =0 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*40}=\frac{8}{80} =1/10 $
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