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(y-1^)2=15+y^2
We move all terms to the left:
(y-1^)2-(15+y^2)=0
We multiply parentheses
-(15+y^2)+2y+=0
We get rid of parentheses
-y^2+2y-15+=0
We add all the numbers together, and all the variables
-1y^2+2y=0
a = -1; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-1)·0
Δ = 4
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{4}=2$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-1}=\frac{-4}{-2} =+2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-1}=\frac{0}{-2} =0 $
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