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5y^2-23y+12=0
a = 5; b = -23; c = +12;
Δ = b2-4ac
Δ = -232-4·5·12
Δ = 289
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{289}=17$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-17}{2*5}=\frac{6}{10} =3/5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+17}{2*5}=\frac{40}{10} =4 $
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