$\displaystyle [(x-2)^2]-9, x is greater than or equal to 4$
$\displaystyle [(x-2)^2]-9 = A$
$\displaystyle x^2-4x-5=A$
$\displaystyle (x+1)(x-5)=A$
that's how far i got...
/edit never mind I figured it out.
$\displaystyle [(x-2)^2]-9, x is greater than or equal to 4$
$\displaystyle [(x-2)^2]-9 = A$
$\displaystyle x^2-4x-5=A$
$\displaystyle (x+1)(x-5)=A$
that's how far i got...
/edit never mind I figured it out.