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3y^2-13y=0
a = 3; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·3·0
Δ = 169
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{169}=13$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*3}=\frac{0}{6} =0 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*3}=\frac{26}{6} =4+1/3 $
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