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factor 2x^2+3x+11
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Its not rational, so pretty much put prime
You're going to want t complete the square with this one
@Kimes are you there?
yeah sorry, but i got x(2x+3)+11
Cannot be factored. A plot is attached.
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The only thing you can do is use the completing the square method to put in a form to solve for x.
Ah okay I got it, thanks
Begin with grouping terms with x in them. \[(2x^2+3x)+11\]\[2\left(x^2+\frac{3}{2}x\right)+11\]\[2\left(x^2+\frac{3}{2}x+\frac{9}{16}\right)+11 - \frac{9}{8}\]\[\boxed{2\left(x+\frac{3}{4}\right)^2+\frac{79}{8}}\]
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