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28y^2+39y+8=0
a = 28; b = 39; c = +8;
Δ = b2-4ac
Δ = 392-4·28·8
Δ = 625
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{625}=25$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-25}{2*28}=\frac{-64}{56} =-1+1/7 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+25}{2*28}=\frac{-14}{56} =-1/4 $
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