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3y^2-28y+60=0
a = 3; b = -28; c = +60;
Δ = b2-4ac
Δ = -282-4·3·60
Δ = 64
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{64}=8$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-8}{2*3}=\frac{20}{6} =3+1/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+8}{2*3}=\frac{36}{6} =6 $
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