proof that quadratic equation has real roots for $p+4\geq0$ (pls help, urgent!!)

Question

How to proof that $0=-x^2 + 4x + p$ has real roots when $p+4\geq0$

I take the discriminant $b^2-4ac$ $=>$ $4^2 -4(-1)(p)$ =$16+4p$

Then I assume $a=16+4p$ $=>$ $a/4=4+p$

Substitute $a/4$ into $p+4\geq0$

I get $a/4\geq0$ then $a\geq0$

Thus, $16+4p\geq0$

$p\geq-4$

What????

Answer 1

What seems to be the problem?
$p\ge-4 \implies p+4\ge -4+4 \implies p+4\ge 0$

You have solved it correctly.
Perhaps you did not read the question thoroughly?

Answer 2

Suppose $p$ is a real number. The discriminant of your polynomial is $D = 16 + 4p$. There is a real root iff $D\geq 0$ which means that $4+p\geq 0$, i.e., $p\geq -4$.