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3y^2-23y+40=0
a = 3; b = -23; c = +40;
Δ = b2-4ac
Δ = -232-4·3·40
Δ = 49
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{49}=7$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-7}{2*3}=\frac{16}{6} =2+2/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+7}{2*3}=\frac{30}{6} =5 $
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