How do you solve the equation $x^2-3x-18=0$ by using the quadratic formula?

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Answer 1 · 2018-05-11T11:18:53

$x=-3 and x=6$

It is really worth trying to commit this formula to memory.

Given: $x^2-3x-18=0$

Compare to $ax^2+bx+c=0$ where $x=(-b+-sqrt(b^2-4ac))/(2a)$

Comparing the two we have:

$a=1; b=-3 and c=-18$

$x=(-(-3)+-sqrt((-3)^2-4(1)(-18)))/(2(1))$

$x=(3+-sqrt(81))/(2(1))$

$x=(3+-9)/2$

$x=-3 and x=6$

How do you solve the equation x^2-3x-18=0x^2-3x-18=0 by using the quadratic formula?