Question

solve x^2+2x+4=0 by completing the square ​
Plz answer fast

Answer 1

${x}^{2} + 2x + 4 = 0\\ divide \: both \: sides \: with \: {x}^{2} \\ coefficient.... \\ as \: the \: coefficient \: is \: one \: the \: equation \: \\ remains \: same \\ {x}^{2} + 2x = - 4 \\ {x}^{2} + 2.1.x = - 4 \\ {x}^{2} + 2.1.x + ( {1}^{2} ) = - 4 + ( {1}^{2} ) \\ to \: turn \: the \: equation \: in \: to \: formula \\ form \: we \: are \: adding \: ( {1}^{2} ) \: on \: both \\ sides \: of \: equation.. \\ now \: this \: is \: in \: the \: form \: of \: \\ {a}^{2} + 2ab + {b}^{2} ..... \\ (a + b) {}^{2} = {a}^{2} + 2ab + {b}^{2} \\ (x + 1) {}^{2} = - 4 + 1 \\ (x + 1) {}^{2} = - 3 \\ x + 1 = + \: or - \sqrt{ - 3} \\ x = + \: or - \sqrt{ - 3} + 1 \\ x = + \sqrt{ - 3} + 1 \\x = \sqrt{ - 3} + 1 \\ also \: x = - \sqrt{ - 3} + 1$

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