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10y^2+43y+28=0
a = 10; b = 43; c = +28;
Δ = b2-4ac
Δ = 432-4·10·28
Δ = 729
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{729}=27$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-27}{2*10}=\frac{-70}{20} =-3+1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+27}{2*10}=\frac{-16}{20} =-4/5 $
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