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28y^2-9y-4=0
a = 28; b = -9; c = -4;
Δ = b2-4ac
Δ = -92-4·28·(-4)
Δ = 529
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{529}=23$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-23}{2*28}=\frac{-14}{56} =-1/4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+23}{2*28}=\frac{32}{56} =4/7 $
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