<s>[INST] Find length (requires quadratic equation) There is an equation $l = 4\sqrt{(\frac{w^2+d^2}{4})+(h-k)^2}$
Following values are given
$l = 50$
$w = 12$
$d = 10$
$k = 3$
What is $h$?
Following choices are
$14$
$10$
$11$
$12$
$13$
My answers come nothing near these!
Here is what I am doing, please assist
TL;DR: $b = 27.75, b=8.245$

$50 = 4\sqrt{(\frac{w^2+d^2}{4})+(h-k)^2}$
$50 = 4\sqrt{(\frac{12^2+10^2}{4})+(h-3)^2}$
$50 = 4\sqrt{(\frac{244}{4})+h^2-6h+9^2}$
$50 = 4\sqrt{61+h^2-6h+9}$
$50 = 4\sqrt{70+h^2-6h}$
Divide both sides by $4$
$12.5 = \sqrt{70+h^2-6h}$
Raise both sides by power of $2$
$156.25 = 70+h^2-6h$
$0 = 70+h^2-6h - 156.25$
$0 = h^2 - 6h - 86.25$
Now, solve for $h$ with quadratic equation
$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\frac{-(-6)\pm\sqrt{(-6)^2-4(-86.25)}}{2}$
$\frac{36\pm\sqrt{36+345}}{2}$
$\frac{36\pm\sqrt{381}}{2}$
$\frac{36 \pm 19.51}{2}$
$b = 27.75, b=8.245$
 [/INST] From $$\frac{-(-6)\pm\sqrt{(-6)^2-4(-86.25)}}{2}$$
to $$\frac{36\pm\sqrt{36+345}}{2}$$
You have written $36$ for $-(-6)$. If you fix that, what you will find is $h_1 = -6.7596$ and $h_2 = 12.76$. I don't know if question wants you to round it up but with given values, these are the values that $h$ can take. 
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