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5y^2+12y-9=0
a = 5; b = 12; c = -9;
Δ = b2-4ac
Δ = 122-4·5·(-9)
Δ = 324
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{324}=18$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-18}{2*5}=\frac{-30}{10} =-3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+18}{2*5}=\frac{6}{10} =3/5 $
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