2x2 + 8x - 3 = 0

My method:

\(\displaystyle 2((x^2+8x+16)-16)-3\longrightarrow 2((x+4)^2-16)-3 \longrightarrow 2(x+4)^2=35\)

\(\displaystyle 2(x+4)(x+4)=35\ is\ 2x^2+16x+32=35\)

which is \(\displaystyle 2x^2+16x-3=0\)

Instead \(\displaystyle 2x^2+8x=3\)

\(\displaystyle x^2+4x=\frac{3}{2}\)

\(\displaystyle x^2+2(2x)+2^2-2^2=\frac{3}{2}\)

\(\displaystyle (x+2)^2-4=\frac{3}{2}\)

\(\displaystyle (x+2)^2-\frac{11}{2}=0\)

\(\displaystyle 2(x+2)^2-11=0\)