⇒x2+2x+(1+4)=0
⇒(x2+2x+1)+4=0
⇒(x+1)2−4(−1)=0
[∵a2+2ab+b2=(a+b)2]
⇒(x+1)2−(4i2)=0[∵i2=−1]
⇒(x+1)2−(4i2)=0
⇒(x+1)2−(2i)2=0
⇒(x+1−2i)(x+1+2i)=0
[∵a2−b2=(a−b)(a+b)]
⇒(x+1−2i)=0or(x+1+2i)=0
⇒x=−1+2iorx=−1−2i
Hence, the roots are −1+2iand−1−2i.