Question

What are the real zeros for the function f(x)=x^2-3x-28

Answer 1

$x^2-3x-28=0\\ x^2-7x+4x-28=0\\ x(x-7)+4(x-7)=0\\ (x+4)(x-7)=0\\ x=-4 \vee x=7$

Answer 2

Factor the equation: (x-7)(x+4) Solve for X: X-7=0 x+4=0 X=7 x=-4 A zero is when the function crosses the x-axis which is solved by setting the function equal to zero.