Question

could you help me with this problem please ​

Answer 1

1. Using the quadratic formula

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

$x=(-4\pm\sqrt{16-64} )/2$

$x=[-4\pm(-4\sqrt{3}i)]/2\\x=-2\pm2\sqrt{3}i$

so, its c

-Hunter

Answer 2

Answer:

$x = -2 \pm 2i\sqrt{3}}$

Explanation:

                      given quadratic sample: x²+4x+16=0

Quadratic formula: $x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{2a}$

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

using the formula:

→ $x = \frac{ -4 \pm \sqrt{4^2 - 4*1*16}}{2*1}$

→ $x = \frac{ -4 \pm \sqrt{-48}}{2}$

→ $x = \frac{ -4 \pm 4i\sqrt{3}}{2}$

→ $x = -2 \pm 2i\sqrt{3}}$