need some guidance with a quadratic equation. Suppose $x^2+20x-4000=0$

Here is what I have done so far; Using the Quadratic Equation $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where from the above equation, $a=1$, $b=20$, $c=-4000$, we find

\begin{align*}x&=\frac{-20+\sqrt{(20)^2-(4)(1)(-4000)}}{2\times 1}\\ &=\frac{-20+\sqrt{16400}}{1}\\ &=108.0624847 \end{align*}

and

\begin{align*} x&=\frac{-20-\sqrt{20^2-(4)(1)(-4000)}}{2\times1}\\ &=\frac{-20-\sqrt{16400}}{1} \\ &=-148.0624847 \end{align*}

Neither of these values for $x$ prove correct when applied to $x^2+20x-4000=0$. I know something is wrong but I can’t figure out what. Any guidance would be appreciated.

• Can you please use Mathjax? – suomynonA Nov 27 '16 at 19:07
• The denominator is correct, but the nominator should be $10(\pm \sqrt{41}-1)$. – Dietrich Burde Nov 27 '16 at 19:20

$2\times 1=2$, not $1$. In the denominator.

• Thanks Tilper, I can't believe I didn't spot that. I feel silly. – Moose Nov 27 '16 at 20:13

Did you figure it out?? I think everyone pointed out where you went wrong, but just to be sure:

$x=\frac{-20+\sqrt{20^2-(4)(1) (-4000)}}{2*\ 1}$

$\frac{1}{2} (-20-20 \sqrt{41})$

= $54.03124237$

And similarly

$x=\frac{-20-\sqrt{20^2-(4)(1) (-4000)}}{2*\ 1}$

=$\frac{1}{2} (-20+20 \sqrt{41})$

$=-74.03124237$

I used Mathematica to check my work:

Letting $answer\text{=}-74.03124237$

$=answer^2+20 answer-4000$

= 0

Similarly,

$answer\text {=} 54.03124237$

$=answer^2+20 answer-4000$

=0