First, we need to transform this into quadratic form by subtracting #color(red)(25)# from each side of the equation to keep the equation balanced while equating to #0#:
#9x^2 - color(red)(25) = 25 - color(red)(25)#
#9x^2 - 25 = 0#
This is a special form of the quadratic equation which has the solution:
#color(red)(a)x^2 - color(blue)(b) = (sqrt(color(red)(a))x + sqrt(color(blue)(b)))(sqrt(color(red)(a))x - sqrt(color(blue)(b))) = 0#
Substituting from our quadratic gives:
#color(red)(9)x^2 - color(blue)(25) = (sqrt(color(red)(9))x + sqrt(color(blue)(25)))(sqrt(color(red)(9))x - sqrt(color(blue)(25))) = 0#
#(3x + 5)(3x - 5) = 0#
or, because the #sqrt(9) = +-3#
#(-3x + 5)(-3x - 5) = 0#
Now, we can solve each term for #0#:
Solution 1)
#3x + 5 = 0#
#3x + 5 - 5 = 0 - 5#
#3x + 0 = -5#
#3x = -5#
#(3x)/3 = -5/3#
#x = -5/3#
Solution 2)
#3x - 5 = 0#
#3x - 5 + 5 = 0 + 5#
#3x + 0 = 5#
#3x = 5#
#(3x)/3 = 5/3#
#x = 5/3#
Solution 3)
#-3x + 5 = 0#
#-3x + 5 - 5 = 0 - 5#
#-3x + 0 = -5#
#-3x = -5#
#(-3x)/(-3) = (-5)/-3#
#x = 5/3#
Solution 4)
#-3x - 5 = 0#
#-3x - 5 + 5 = 0 + 5#
#-3x + 0 = 5#
#-3x = 5#
#(-3x)/-3 = 5/-3#
#x = -5/3#
#x = 5/3# or #x = -5/3#
