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2y^2=17y
We move all terms to the left:
2y^2-(17y)=0
a = 2; b = -17; c = 0;
Δ = b2-4ac
Δ = -172-4·2·0
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-17}{2*2}=\frac{0}{4} =0 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+17}{2*2}=\frac{34}{4} =8+1/2 $
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