Question

Solve −3x^2 − 4x − 4 = 0.

Answer 1

Must use quadratic formula (basically completing the square)

so

for an equation
ax^2+bx+c=0
x=$\frac{-b+/- \sqrt{b^2-4ac} }{2a}$

ax^2+bx+c=0
-3x^2-4x-4=0
a=-3
b=-4
c=-4

x=$\frac{-(-4)+/- \sqrt{(-4)^2-4(-3)(-4)} }{2(-4)}$
x=$\frac{4+/- \sqrt{16-48} }{-8}$
x=$\frac{4+/- \sqrt{-32} }{-8}$
x=$\frac{4+/- (\sqrt{-1})(\sqrt{32}) }{-8}$
x=$\frac{4+/- (i)(\sqrt{(4)(4)(2)}) }{-8}$
x=$\frac{4+/- (i)(4\sqrt{2}) }{-8}$
x=$\frac{4+/- 4i\sqrt{2} }{-8}$
x=$\frac{1+/- i\sqrt{2} }{-2}$

x=$\frac{1+ i\sqrt{2} }{-2}$ or $\frac{1- i\sqrt{2} }{-2}$
or
x=$\frac{-1- i\sqrt{2} }{2}$ or $\frac{-1+ i\sqrt{2} }{2}$

Answer 2

Answer:

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

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

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

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

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

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