How do you solve $x^2 - 3x - 18 = 0$ ?

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Answer 1 · 2015-11-02T21:19:09

$x_1=6$ and $x_2=-3$

you have an Equation like this: $ax^2+bx+c=0$

Then you can use the formula for solving quadratic Equations:
$x=(-b+-sqrt(b^2-4ac))/(2a)$

so we have: $x=-(-3)(+-)sqrt((-3)^2-4*1*(-18))/(2*1)$

$x=(3+-sqrt(9+72))/2=(3+-9)/2$

$x_1=(3+9)/2=6$
$x_2=(3-9)/2=-3$

How do you solve x^2 - 3x - 18 = 0x^2 - 3x - 18 = 0?