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15y^2+47y+28=0
a = 15; b = 47; c = +28;
Δ = b2-4ac
Δ = 472-4·15·28
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(47)-23}{2*15}=\frac{-70}{30} =-2+1/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+23}{2*15}=\frac{-24}{30} =-4/5 $
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