Which equation has two real solutions? A x^2-4=0 B x-4=0 C x^2+4=0 D x+4=0

To determine which equation has two real solutions, we need to analyze each equation individually.<br /><br />A. $x^{2}-4=0$<br />This equation can be factored as $(x-2)(x+2)=0$. Therefore, the solutions are $x=2$ and $x=-2$. These are two real solutions.<br /><br />B. $x-4=0$<br />This equation has only one solution, which is $x=4$.<br /><br />C. $x^{2}+4=0$<br />This equation can be rewritten as $x^{2}=-4$. Since the square of a real number cannot be negative, this equation has no real solutions.<br /><br />D. $x+4=0$<br />This equation has only one solution, which is $x=-4$.<br /><br />Therefore, the equation that has two real solutions is option A: $x^{2}-4=0$.