\(x^{2} - 3x - 18 = 0\) factors → \((x - 6) * (x+ 3)\)
If you want to find your roots, use Bhaskara's formula
\(x^{2} - 3x - 18 = 0\)
a = 1; b = - 3, c = - 18
\(\Delta = b^2 - 4*a*c\)
\(\Delta = (-3)^2 - 4*1*(-18)\)
\(\Delta = 9 + 72\)
\(\Delta = 81\)
\(x = \frac{-b\pm \sqrt{\Delta} }{2*a}\)
\(x = \frac{-(-3)\pm \sqrt{81} }{2*(-3)}\)
\(x = \frac{3\pm 9 }{-6}\)
\(x' = \frac{3-9}{-6} = \frac{-6}{-6} \rightarrow \boxed{x' = 1}\)
\(x'' = \frac{3+9}{-6} = \frac{12}{-6} \rightarrow \boxed{x'' = -2}\)
But, the answer to your question are the factors of the equation, here is:
ANSWER:
\((x - 6) * (x+ 3)\)
x - 6 and x + 3
