Verified Answer :

apologiabiology

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 \(\frac{-1+ i\sqrt{2} }{2}\)

calculista

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

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

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

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



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