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3y^2-32y+69=0
a = 3; b = -32; c = +69;
Δ = b2-4ac
Δ = -322-4·3·69
Δ = 196
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{196}=14$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-14}{2*3}=\frac{18}{6} =3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+14}{2*3}=\frac{46}{6} =7+2/3 $
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