Solve the equation:`x^2+3x+9=0`

Explanation

The given equation is `x^(2)+3x+9=0`.
This is of the form `ax^(2)+bx+c=0`, where a = 1, b = 3 and c = 9.
`therefore" "(b^(2)-4ac)=(3^(2)-4xx1xx9)=(9-36)=27 lt 0`.
So, the given equation has complex roots.
These roots are given by
`(-b+- sqrt(b^(2)-4ac))/(2a)=(-3 +- sqrt(-27))/(2xx1)" "[because b^(2)-4ac = -27]`
`=(-3 +- isqrt(27))/(2)=(-3 +- i3 sqrt(3))/(2)`
`therefore" ""solution set"={(-3+ i3 sqrt(3))/(2),(-3-i3 sqrt(3))/(2)}={(-3)/(2)+(3sqrt(3))/(2)i, (-3)/(2)-(3sqrt(3))/(2)i}`.