Question

how would you solve c^2+9c+18=0 because i think its non-factorable?

Answer 1

there are 2 methods

1 completing the square

which requires rewriting the equation

\[c^2 + 9c = -18\]

then finding the value of (9/2)^2 and adding it to both sides

\[c^2 + 9c + \frac{81}{4} = -18 + \frac{81}{4}\]

left hand side is a perfect square and the RHS is just a number


2 using the general quadratic formula

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Answer 2

thanks dude i forgot about the general quadratic formula its easier for me to use that thanks