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-3y^2+25y-28=0
a = -3; b = 25; c = -28;
Δ = b2-4ac
Δ = 252-4·(-3)·(-28)
Δ = 289
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{289}=17$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-17}{2*-3}=\frac{-42}{-6} =+7 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+17}{2*-3}=\frac{-8}{-6} =1+1/3 $
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