Question

Could you help me with this problem please ​

Answer 1

1. Using the quadratic formula

`x=(-b\pm\sqrt{b^(2)-4ac})/2a`

`x=(-4\pm√(16-64) )/2`

`x=[-4\pm(-4√(3)i)]/2\\x=-2\pm2√(3)i`

so, its c

-Hunter

Answer 2

`x = -2 \pm 2i√(3)}`

Step-by-step explanation:

given quadratic sample: x²+4x+16=0

Quadratic formula:
`x = ( -b \pm √(b^2 - 4ac))/(2a)`

Here a = 1, b = 4, c = 16 [ which are coefficients from ax²+bx+c ]

using the formula:

→
`x = ( -4 \pm √(4^2 - 4*1*16))/(2*1)`

→
`x = ( -4 \pm √(-48))/(2)`

→
`x = ( -4 \pm 4i√(3))/(2)`

→
`x = -2 \pm 2i√(3)}`