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y=5/3y+2
We move all terms to the left:
y-(5/3y+2)=0
Domain of the equation: 3y+2)!=0We get rid of parentheses
y∈R
y-5/3y-2=0
We multiply all the terms by the denominator
y*3y-2*3y-5=0
Wy multiply elements
3y^2-6y-5=0
a = 3; b = -6; c = -5;
Δ = b2-4ac
Δ = -62-4·3·(-5)
Δ = 96
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4\sqrt{6}}{2*3}=\frac{6-4\sqrt{6}}{6} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4\sqrt{6}}{2*3}=\frac{6+4\sqrt{6}}{6} $
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