Simplifying Complex Quadratic Equations

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Homework Statement



x^2 + (1-i)x + (-6 + 2i) = 0, solve in terms of a + bi

Homework Equations



The Attempt at a Solution



Here's what I have so far... But I could be wrong.

x = -(x-xi) +/- sqrt[ (x-xi)^2 - 4(-6 + 2i) ] / 2

x = -(x-xi) +/- sqrt[ (x^2 - 2xi + xi^2) + 24 - 8i ] / 2

I'm having trouble simplifying this part. Did I do something wrong?
 
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Remember that x should not be on the right hand side of the equation; b is just 1-i, not (1-i)*x.
 
Ohhhh snap. Let me try again.
 
So now I get to

-(1 - i) +/- sqrt[ 25 - 10i + i^2 ] / 2

And again, I'm stuck on the simplification.
 
Check for factors, you have another quadratic under the root
 
So I factor that to (i - 5)^2 and get:

-(1 - i) +/- (i -5) / 2...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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