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3y^2-13y+9=0
a = 3; b = -13; c = +9;
Δ = b2-4ac
Δ = -132-4·3·9
Δ = 61
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{-(-13)-\sqrt{61}}{2*3}=\frac{13-\sqrt{61}}{6} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{61}}{2*3}=\frac{13+\sqrt{61}}{6} $
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