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5y+y^2=1
We move all terms to the left:
5y+y^2-(1)=0
a = 1; b = 5; c = -1;
Δ = b2-4ac
Δ = 52-4·1·(-1)
Δ = 29
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{-(5)-\sqrt{29}}{2*1}=\frac{-5-\sqrt{29}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{29}}{2*1}=\frac{-5+\sqrt{29}}{2} $
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