Question 147540
{{{16y^2=25}}}. I am assuming that you have to solve for y. First, you factor. First, move 25 over to the other side so that it is all set equal to zero (since y is sqaured, you must do this).  {{{16y^2-25=0}}}.  Now you see that {{{16y^2-25}}} is the difference between two perfect squares.  so factor it out as the difference of the square roots times the sum of the square roots. {{{(4y-5)(4y+5)}}}.  That is how to factor the original equation. Then plug in the values of y that would make each expression (individually) zero, because it is set equal to zero.  Ask yourself, what minus five would give me zero? You got five right? Good.  Now ask yourself, four times what would give me five? You should have gotten 5/4.  Now let's move to the other expression.  This time, ask, what plus five would give me zero? You should get -5.  Now ask, 4 times what would give me -5? You should get -5/4.  These are your two values for y. There are two values because in the original equation, y is squared. So y=5/4;-5/4.