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26y^2-51y+25=0
a = 26; b = -51; c = +25;
Δ = b2-4ac
Δ = -512-4·26·25
Δ = 1
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{1}=1$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-51)-1}{2*26}=\frac{50}{52} =25/26 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-51)+1}{2*26}=\frac{52}{52} =1 $
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