Question 223959
{{{25x^2-16=0}}}
Can be rewritten as:
{{{(5x)^2- 4^2=0}}}
Now, it is a "difference of squares" -- which is a special factor case:
{{{(5x+4)(5x-4)=0}}}
To solve, set each to zero -- which yields:
x = {-4/5 , 4/5}