Answer:
no real solution
Step-by-step explanation:
[tex]-3x^2-4x-4=0\qquad\text{change the signs}\\\\3x^2+4x+4=0\\\\\text{use the quadratic formula:}\\\\\text{for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac<0,\ \text{then an equation has no solution}\\\\\text{if}\ b^2-4ac=0,\ \text{then an equation has one solution}\ x=\dfrac{-b}{2a}\\\\\text{if}\ b^2-4ac>0,\ \text{then an equation has two solutions:}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]\text{We have}\ a=3,\ b=4,\ c=4.\\\\\text{Substitute:}\\\\b^2-4ac=4^2-4(3)(4)=16-48=-32<0\\\\\bold{no\ real\ solution}[/tex]
