We know for our problem that the zeroes of our quadratic equation are [tex](2-3i)[/tex] and [tex](2+3i)[/tex], which means that the solutions for our equation are [tex]x=2-3i[/tex] and [tex]x=2+3i[/tex]. We are going to use those solutions to express our quadratic equation in the form [tex]a x^{2} +bx+c[/tex]; to do that we will use the zero factor property in reverse: [tex]x=2-3i[/tex] [tex]x-2=-3i[/tex] [tex]x-2+3i=0[/tex] [tex]x=2+3i[/tex] [tex]x-2=3i[/tex] [tex]x-2-3i=0[/tex] Now, we can multiply the left sides of our equations: [tex](x-2+3i)(x-2-3i)= x^{2} -2x-3ix-2x+4+6i+3ix-6i-3^2i^2[/tex] = [tex] x^{2}-4x+4-9i^{2}[/tex] = [tex] x^{2} -4x+4+9[/tex] = [tex] x^{2} -4x+13[/tex] Now that we have our quadratic in the form [tex]a x^{2} +bx+c[/tex], we can infer that [tex]a=1[/tex] and [tex]b=-4[/tex]; therefore, we can conclude that [tex]a+b=1+(-4)=1-4=-3[/tex].
