(y+3)2+y2=9

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Solution for (y+3)2+y2=9 equation:



(y+3)2+y2=9
We move all terms to the left:
(y+3)2+y2-(9)=0
We add all the numbers together, and all the variables
y^2+(y+3)2-9=0
We multiply parentheses
y^2+2y+6-9=0
We add all the numbers together, and all the variables
y^2+2y-3=0
a = 1; b = 2; c = -3;
Δ = b2-4ac
Δ = 22-4·1·(-3)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4}{2*1}=\frac{-6}{2} =-3 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4}{2*1}=\frac{2}{2} =1 $

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