How do you find all the real and complex roots of $18x²+3x-1=0$ ?

Question

1696 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-01T11:56:43

here is how,

as $x$ has the highest power of $2$ , there are two values of $x$

we know,

$x=(-b+-sqrt(b^2-4ac))/(2a)$

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

$=(-3+-sqrt(9+72))/36$

$=(-3+-sqrt81)/36$

$=(-3+-9)/36$

using the + value,

$x=(-3+9)/36$

$=6/36$

$=1/6$

using the - value,
$x=(-3-9)/36$

$=-12/36$

$=-1/3$

How do you find all the real and complex roots of 18x²+3x-1=018x²+3x-1=0?