(3-y)2+y2=9

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

(3-y)2+y2=9

We move all terms to the left:

(3-y)2+y2-(9)=0

We add all the numbers together, and all the variables

(-1y+3)2+y2-9=0

We add all the numbers together, and all the variables

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