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3y^2-17y+24=0
a = 3; b = -17; c = +24;
Δ = b2-4ac
Δ = -172-4·3·24
Δ = 1
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{1}=1$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-1}{2*3}=\frac{16}{6} =2+2/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+1}{2*3}=\frac{18}{6} =3 $
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