Answer:
x equals quantity of negative \(2\) plus or minus \(2i\)i 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
\(x=\frac{-2(+/-)2i\sqrt{2}} {3}\)
\(x1=\frac{-2(+)2i\sqrt{2}} {3}\)
\(x2=\frac{-2(-)2i\sqrt{2}} {3}\)
