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135y^2+498y+391=0
a = 135; b = 498; c = +391;
Δ = b2-4ac
Δ = 4982-4·135·391
Δ = 36864
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{36864}=192$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(498)-192}{2*135}=\frac{-690}{270} =-2+5/9 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(498)+192}{2*135}=\frac{-306}{270} =-1+2/15 $
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