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(y-11)3y=1
We move all terms to the left:
(y-11)3y-(1)=0
We multiply parentheses
3y^2-33y-1=0
a = 3; b = -33; c = -1;
Δ = b2-4ac
Δ = -332-4·3·(-1)
Δ = 1101
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{-(-33)-\sqrt{1101}}{2*3}=\frac{33-\sqrt{1101}}{6} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+\sqrt{1101}}{2*3}=\frac{33+\sqrt{1101}}{6} $
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