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2y=2/5y+3
We move all terms to the left:
2y-(2/5y+3)=0
Domain of the equation: 5y+3)!=0We get rid of parentheses
y∈R
2y-2/5y-3=0
We multiply all the terms by the denominator
2y*5y-3*5y-2=0
Wy multiply elements
10y^2-15y-2=0
a = 10; b = -15; c = -2;
Δ = b2-4ac
Δ = -152-4·10·(-2)
Δ = 305
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{305}}{2*10}=\frac{15-\sqrt{305}}{20} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{305}}{2*10}=\frac{15+\sqrt{305}}{20} $
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