How do you solve $y = 2x^2 + 3x + 6$ ?

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Answer 1 · 2015-10-27T19:07:12

Assuming you mean:
What are the roots? (find x if y=0)
It does not cross the x-axis so has a complex number solution.
$c = 1/4(3 +-(sqrt(39))i)$

Tony B Tony B

Using standard form equations of:

$y =ax^2 + bx + c " "$ and $" x = (-b+-sqrt(b^2 - 4ac))/(2a)$

Where $a=2; " " b = 3; " and " c =6$ giving:

$x = (-3 +- sqrt( (3)^2 - 4(2)(6)))/(2(2)$

$x= (-3 +- sqrt( 9 - 48))/4$

$x=-3/4 +- sqrt((-39))/4$

But $-39 = 39 times (-1)$
and $sqrt(-1) = i$

giving: $x = 1/4(3 +-(sqrt(39))i)$

How do you solve y = 2x^2 + 3x + 6y = 2x^2 + 3x + 6?