Question

PLEASE HELP

Answer 1

Answer: The correct solution is option third.

Explanation:

The given equation is,

$-3x^2-4x-4=0$

The quadratic formula to find the value of x for the equation  $ax^2+bx+c=0$  is,

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

The values are,

$a=-3,b=-4,c=-4$

$x=\frac{-(-4)\pm \sqrt{(-4)^2-4(-3)(-4)}}{2(-3)}$

$x=\frac{4\pm \sqrt{16-48}}{-6}$

$x=\frac{4\pm \sqrt{-32}}{-6}$

Since $\sqrt{-1}=i$,

$x=\frac{4\pm 4i\sqrt{2}}{-6}$

$x=\frac{2(2\pm 2i\sqrt{2})}{-6}$

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

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

Therefore third option is correct.

Answer 2

Answer:

x equals quantity of negative $2$ plus or minus $2i$ square root of $2$ all over $3$

Step-by-step explanation:

we have

$-3x^{2} -4x-4=0$

Rewrite (Multiply by $-1$ both sides)

$3x^{2}+4x+4=0$

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

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

in this problem we have

$3x^{2}+4x+4=0$

so

$a=3\\b=4\\c=4$

substitute

$x=\frac{-4(+/-)\sqrt{4^{2}-4(3)(4)}} {2(3)}$

$x=\frac{-4(+/-)\sqrt{-32}} {6}$

remember that

$i=\sqrt{-1}$

$x=\frac{-4(+/-)4i\sqrt{2}} {6}$

Simplify

$x=\frac{-2(+/-)2i\sqrt{2}} {3}$

$x1=\frac{-2(+)2i\sqrt{2}} {3}$

$x2=\frac{-2(-)2i\sqrt{2}} {3}$