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3y^2-16y+20=0
a = 3; b = -16; c = +20;
Δ = b2-4ac
Δ = -162-4·3·20
Δ = 16
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{16}=4$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4}{2*3}=\frac{12}{6} =2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4}{2*3}=\frac{20}{6} =3+1/3 $
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