How to solve this equation

x² - 7x + 12 = 0

\(x^{2}-7x+12=0\)

\(x^{2}-4x-3x+12=0\)

x(x-4)-3(x-4)=0

(x-3)(x-4)=0

Equating each term to 0

x-3=0

x=3

x-4=0

x=4

x=3 and 4