<s>[INST] Find the equation whose roots are each six more than the roots of $x^2 + 8x - 1 = 0$ 
Find the equation whose roots are each six more than the roots of $x^2 + 8x - 1 = 0$

I must use Vieta's formulas in my solution since that is the lesson we are covering with our teacher.
My solution:
Let p and q be the roots of the quadratic.
$$\begin{align} p + q = & -8 \\ pq = & -1 \end{align}$$
If the roots are each six more than the roots of the quadratic, then we will have:
$$\begin{align} p + q + 12 = & -8 \\ 
p + q = & -20 \\
pq = & - 3 \end{align}$$
Also, 
$$ \begin{align} (p+6)(q+6) = & pq + 6p + 6q + 36\\
=&pq + 6(p+q) + 36 \\
=&-3 + 6(-8) + 36 \\
=&-3 - 48 + 36 \\
&36 - 51 = -15 \end{align} $$
Hence, -15 = constant.
Thus, the quadratic equation is  $x^2 + 20x - 15$
My worksheet gives an answer of $x^2 - 4x - 13$, how am I wrong?
Thanks!
 [/INST] Let $x_1,x_2$ be the roots of the equation: $x^2+8x-1 = 0$, and let $y_1 = x_1 +6, y_2 = x_2+6 \Rightarrow y_1+y_2 = x_1+x_2+12=-8+12 = 4, y_1y_2 = (x_1+6)(x_2+6)=x_1x_2+6(x_1+x_2)+36=-1+6\cdot (-8) + 36=-13\Rightarrow y^2-4y-13=0$ is the sought after equation.
</s>