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5y^-3y^2+8y=0
We add all the numbers together, and all the variables
-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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*-3}=\frac{-26}{-6} =4+1/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*-3}=\frac{0}{-6} =0 $
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