What indicators predict that a quadratic funtion will have a complex solution? [Solve and show all work] x^2 - 10x = -13 x = -b +- squareroot b^2 -4ac OVER 2a

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What indicators predict that a quadratic funtion will have a complex solution? [Solve and show all work] x^2 - 10x = -13 x = -b +- squareroot b^2 -4ac OVER 2a

Mathematics
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focus on "b^2 -4ac " as you need to take its square root.....
Well, I'm not sure how to solve it. Can you teach me? @IrishBoy123
what is b and what is c in your original quadratic?

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this is the quadratic you posted: \(\large x^2 - 10x = -13 \) and this is the additional formula you also posted: \(\large x = \frac{-b \ \pm \ \sqrt{ b^2 -4ac} }{ 2a}\)
maybe it would be better to start by asking you what you are trying to do to the quadratic? i know but do you also know?
No, I'm unsure on how to solve this question...
that's my question....you are trying to find the value of x where the quadratic = 0, is that right?
Well, the question is asking to find complex solutions... Which I dont really understand what it means by that.
and I guess it's asking to solve also.. So yeah. solving for x
take your original equation \(x^2 - 10x = -13 \) and re-write it as \(x^2 - 10x +13 =0\) , yep?!! then compare it to this \(a x^2 + b x + c = 0\) and then apply that second equation you posted, the one with "\(b^2 - 4 ac\)" in it [i set it out again above] that second equation gives you the roots (x intercepts) of the quadratic, ie x values for when quadratic = 0 try it and see where you go.....
clearly if \(b^2 - 4ac \lt 0\) you might have an issue (ie complex roots) but run with it first....
Well, I'm sorta confused on how to plug it in and solve it... Can you help me on that?
@IrishBoy123 you there?
compare \(x^2−10x+13=0\) to \(ax^2+bx+c=0\) so \(a = 1, b = -10, c = 13\) plug these into \(\huge x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
what do you get?
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okay, I regret not paying enough attention in math class... How do I solve this? Sorry to take up more of your time by asking this. I never got anything out of quadratics... *sigh*
cool, but its \(-b \pm\) and -(-10) = 10 but now get out your calculator/spreadsheet/whatever and calculate the 2 values of x. it is \(\pm\) so there will be 2 values. these are your roots
Okay.
\(\huge \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(13)}}{2(1)} \implies \) \(\huge \frac{10 \pm \sqrt{100 - 52}}{2} = 1.54, \ 8.46 \)
yep?
Yeah. Thank you.

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