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33y^2+25y+2=0
a = 33; b = 25; c = +2;
Δ = b2-4ac
Δ = 252-4·33·2
Δ = 361
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{361}=19$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-19}{2*33}=\frac{-44}{66} =-2/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+19}{2*33}=\frac{-6}{66} =-1/11 $
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