First we want to the quadratic trinomial
x2−3x−28. x2−3x−28
x2+4x−7x−28
x(x+4)−7(x+4)
(x−7)(x+4)
We can now use this to write the given equation as
(x−7)(x+4)=0. To solve this equation, we will apply the .
(x−7)(x+4)=0
x−7=0x+4=0(I)(II)
x=7x+4=0
x1=7x2=-4