Using the "zero property" $\displaystyle ax^2+bx+c=0$

$\displaystyle 4x^2=60-x$

move the $\displaystyle 60-x$ over to the left...

$\displaystyle 4x^2-60+x=0$

Factor out...

ok now here is where I have my problem. As seeing that the Eq has anegativeand apositivethe signs would be the same when factoring out no?!

$\displaystyle (4x-15)(x-4)$

$\displaystyle 4x-15=0$ and $\displaystyle x-4=0$

$\displaystyle 4x=+15$ and $\displaystyle x=4$

Thus...

$\displaystyle x=\frac{15}{4}$ and $\displaystyle x=4$

Yes?!