what is the answer and working of x2+y2=5 2y=x-3

$$\\x^2+y^2=5 \qquad(1a)\\
2y=x-3\qquad (2a)\rightarrow \;\;x=2y+3\qquad (2b)\\\\
sub\;2b\;into\;1a\\
\begin{array}{rll}
(2y+3)^2+y^2&=&5\\
4y^2+12y+9+y^2&=&5\\
5y^2+12y+4&=&0\\
5y^2+10y+2y+4&=&0\\
5y(y+2)+2(y+2)&=&0\\
(5y+2)(y+2)&=&0\\
\end{array}\\
y=-2/5\quad or\quad y=-2\\
When\;y=-2\;x=-4+3=-1\\
When\;y=-2/5\;x=-4/5+3=2\frac{1}{5}\\\\
$so the points of intersection are $ (2\frac{1}{5},\frac{-2}{5})\quad and \quad(-1,-2)$$

 

Here are the graphs.

https://www.desmos.com/calculator/bb8eqvm5rz