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For some reason I always get the wrong answer and I don't understand why:

$$4z^2-12z+19=0$$

I got $z= \frac{12 \pm \sqrt{260}i}{8} $ and the answer is supposed to be $z=\frac{3 \pm \sqrt{10}i}{2} $ where is my mistake?

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    $\begingroup$ They are the same thing, only a factor of $4$ was cancelled. $\endgroup$ – Matti P. Jan 17 '19 at 14:14
  • $\begingroup$ Note that $ \sqrt{260} $ (is that is what you meant) is typeset as \sqrt{260} . $\endgroup$ – Martin R Jan 17 '19 at 14:26
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Your mistake is that the discriminant should be $12^2 - 4\times 4\times 19 = -160$, not $-260$, and then dividing top and bottom by $4$ gives the correct answer.

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    $\begingroup$ In applying the quadratic formula … it might help to divide all coefficients by the original a of 4 so that the a drops out or the adjusted a=1 and the adjusted b=-3 and adjusted c=19/4 ... although it's best to keep it as simple as possible in order to minimize the likelihood of adding a mistake caused by extra cleverness … it's useful to understand the root cause of the error, eg b^2=44? rather 144 which lead to the -260 … in order to eliminate the likely missteps in one's problem-solving process. $\endgroup$ – markbruns Jan 18 '19 at 3:49
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When the $x$ coefficient is even, say $ax^2+2bx+c=0,$ there is an equivalent version of the quadratic equation which automatically cancels that extra factor of $4.$

Instead of the usual discriminant $D=b^2-4ac$ with this setup one calculates what might be called the "other" discriminant (for want of a better word) $E=b^2-ac$ as the thing under the radical. As usual if that's negative the complex number $i$ is used when square root of discriminant is taken.

Then the roots are $$\frac{-b \pm \sqrt{E}}{a}.$$

For the re-written example you have: $4z^2-2(-6)z+19=0,$ we get $E=(-6)^2-4\cdot 19=36-4\cdot 19=36-76=-40.$ Then roots are $(6 \pm \sqrt{-40})/12.$ This simplifies again in this case.

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    $\begingroup$ When you write "$ax^2+2bx+c=0$" and "$D=b^2-4ac$", those are two different $b$s. You should pick another letter for the first one. $\endgroup$ – A. Howells Jan 17 '19 at 18:38
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    $\begingroup$ @A.Howells That is clear from my explanation, and is why $E$ was used for $b^2-ac$ instead of the traditional $D$ for the discriminant (when no built-in multiplier of $2$ before $x$-coefficient. $\endgroup$ – coffeemath Jan 17 '19 at 18:49
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    $\begingroup$ @coffeemath It's not clear. The traditional discriminant would be $D = (2b)^2 - 4ac$. $\endgroup$ – JiK Jan 17 '19 at 19:51
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    $\begingroup$ @JiK And that, divided by $4,$ is $b^2-ac.$ $\endgroup$ – coffeemath Jan 18 '19 at 0:20
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    $\begingroup$ @coffeemath Yes, but this answer claims that $D = b^2 - 4ac$, which is wrong. The coefficient of the first degree term is $2b$ here, not $b$. $\endgroup$ – JiK Jan 18 '19 at 1:53

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