Answer:
There are no real number solutions.
Step-by-step explanation:
[tex]2x^{2} +8=0\\2x^{2} =-8\\x^{2} =-4\\x=\sqrt{-4}[/tex]
Real numbers are a whole range of numbers including: Natural Numbers(1;2;3...), Whole Numbers(0;1;2...), Integers(...;-2;-1;0;1;2;...), Rational Numbers(-0.6; -0.2; 0 ;1.5 etc.) and Irrational Numbers([tex]\sqrt{2}[/tex];[tex]\sqrt{3}[/tex];π; etc.). All these number comprise Real Numbers.
Imaginary Numbers are the square roots of negative Integers, which is the case for this question.