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-y^2+2y+1=-1
We move all terms to the left:
-y^2+2y+1-(-1)=0
We add all the numbers together, and all the variables
-1y^2+2y+2=0
a = -1; b = 2; c = +2;
Δ = b2-4ac
Δ = 22-4·(-1)·2
Δ = 12
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{3}}{2*-1}=\frac{-2-2\sqrt{3}}{-2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{3}}{2*-1}=\frac{-2+2\sqrt{3}}{-2} $
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