Q.Solve by the method of perfect squares: $x^2 + 2x - 5 = 0$.

This doesn't factor with small integers, so complete the square directly.

$x^2+2x=5 \Rightarrow x^2+2x+1=6 \Rightarrow (x+1)^2=6 \Rightarrow x=-1\pm\sqrt6$.

Since $a=1$, move the constant across: $x^2+2x=5$.

Half of $2$ is $1$; its square is $1$. Add $1$ to both sides: $x^2+2x+1=5+1=6$.

Left side is a perfect square: $(x+1)^2=6$.

Take square roots: $x+1=\pm\sqrt6$, so $x=-1\pm\sqrt6$.

Check via discriminant: $D=2^2-4(1)(-5)=4+20=24$; $\sqrt{24}=2\sqrt6$; Sridharacharya's formula gives $x=\dfrac{-2\pm2\sqrt6}{2}=-1\pm\sqrt6$ — matches exactly.

Alternative — Sridharacharya's formula directly: $D=4+20=24$, $x=\dfrac{-2\pm\sqrt{24}}{2}=\dfrac{-2\pm2\sqrt6}{2}=-1\pm\sqrt6$.

A frequent slip is computing half of $b=2$ correctly as $1$ but then squaring the ORIGINAL $b$ instead of the halved value (adding $4$ instead of $1$) — always halve first, then square, to get the correct number to add to both sides.