The given quadratic equation is x2+x+1√2=0
This equation can also be written as √2x2+√2x+1=0
On comparing the given equation with ax2+bx+c=0,
we obtain a=√2,b=√2, and c=1
Therefore, the discriminant D=b2−4ac=(√2)2−4(√2)×1=2−4√2
Therefore, the required solutions are
−b±√D2a=−√2±√2−4√22×√2=−√2±√2(1−2√2)2√2
=⎛⎜
⎜⎝−√2±√2(√2√2−1i)2√2⎞⎟
⎟⎠
=−1±1(√2√2−1)i2