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40y^2+9y-9=0
a = 40; b = 9; c = -9;
Δ = b2-4ac
Δ = 92-4·40·(-9)
Δ = 1521
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{1521}=39$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-39}{2*40}=\frac{-48}{80} =-3/5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+39}{2*40}=\frac{30}{80} =3/8 $
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