Question

Find the roots: x2 + 4x + 5 = 0

Answer 1

need to find = x

Given equation- x^2+4x+5=0

By quadratic formula

x^2+4x+5

D=b^2-4ac

D=(4)^2-4(1)(5)

=16-20

D=-4

So it is less than 4 so no real root exist in given equation

and Ty for free points

Answer 2

Answer:

$= > x {}^{2} + 4x + 5 = 0 \\ here \: a = 1 \: b = 4 \: c = 5 \\ x = - b + - \sqrt{b {}^{2} - 4ac }/2a \\ = > - 4 + - \sqrt{(4) {}^{2} - 4(1)(5) }/2(1) \\ = > - 4 + - \sqrt{16 - 20}/2 \\ = > - 4 + - \sqrt{ - 4} /2 \\ = > - 4 + - \sqrt{4} i/2(since = i {}^{2} = - 1) \\ = > - 4 + - 2i/2 \\ = > - 2+ - i \\ x = - 2 + i(or)x = - 2 - i$