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y^2-15.65y+50=0
a = 1; b = -15.65; c = +50;
Δ = b2-4ac
Δ = -15.652-4·1·50
Δ = 44.9225
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15.65)-\sqrt{44.9225}}{2*1}=\frac{15.65-\sqrt{44.9225}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15.65)+\sqrt{44.9225}}{2*1}=\frac{15.65+\sqrt{44.9225}}{2} $
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