A solution is prepared by dissolving 0.56g of benzoic acid (C6H5CO2H), Ka=6,4*10^-5 in enough water to make 1,?

Molar mass C6H5COOH = 122.122 g/mol
0.56g = 0.56/122.122 = 0.00459 mol

This is dissolved in 1 liutre solution:
concentration:
C6H5COOH = 0.00459M

Calculate the [H+] from Ka equation

Ka = [H+] [C6H5COO- ] / [C6H5COOH]
We know the following: [H+] = [C6H5COO-] and the dissociation of the acid is so small that we can call [C6H5COOH] = 0.00459M
Note - doing this avoids the complexity of a quadratic equation - and for such weak acids as benzoic acid, the final result is unchanged
Substitute:
6.4*10^-5 = [H+]² / 0.00459
[H+]² = ( 6.4*10^-5) * 0.00459
[H+]² = 2.9376 *10^-7
[H+] = 5.42*10^-4

Calculate [OH-]
[H+] [OH-] = 10^-14
[OH-] = 10^-14 / ( 5.42*10^-4)
[OH-] = 1.85*10^-11

calculate pH
pH = -log [H+]
pH = -log 5.42*10^-4
pH = 3.27

I used your equation: x² + (6.4*10^-5)x - 2.92*10^-7 = 0 , and solved for X using a quadratic solver which returned a value of [H+] = 5.1*10^-4 This will give pH value = 3.29. This is generally accepted as 3.27 being close enough to warrant the avoidance of the quadratic step. As far as I know most examining bodies accept this, unless they specify that the short cut is not to be used. The idea is that we are learning chemistry here - not mathematics.