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Answer 1 · 2018-06-01T10:10:25

a= 5 and a = -5

Using the difference of two squares:
$a^2-b^2 = (a+b)(a-b)$

$a^2-25 = 0 => (a+5)(a-5)=0$
Using null factor law.
$a+5=0$
$a = -5$ .
Or
$a-5=0$
$a = 5$ .

Answer 2 · 2018-06-01T10:11:06

Is an incomplete second degree equation. See below for solution

We can apply the quadratic formula also but in this case is easy

$a^2-25=0$ we can write

$a^2=25$ (square root in both sides)

$a=+-sqrt25=+-5$

The solutions are $a=5$ and $a=-5$