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!

The original quadratic equation is given to you as $x^2 + 8x - 1 = 0$ . Let it's roots be $(x_1 , x_2)$ . The roots of required equation be $(\alpha ,\beta)$ (say).
Now, $$x_1 + x_2 = -8 \\ x_1 x_2 = -1$$
And $$\alpha = x_1 + 6 \\ \beta = x_2 + 6 $$
So, $$\alpha + \beta = x_1 + x_2 + 12 \\ \alpha \beta = (x_1 + 6)(x_2 + 6) = x_1 x_2 + 6(x_1 + x_2) + 36 $$
You already know the values of $\color{red}{x_1 x_2 = -1} ; \ \color{blue}{x_1 + x_2 = -8}$ . So, just plug-in the values in the above formula. 
$$\begin{align} 
&\alpha + \beta = \color{blue}{x_1 + x_2} + 12 = \color{blue}{-8} + 12 = 4 \\
&\alpha \times \beta = \color{red}{x_1 x_2} + 6\color{Blue}{(x_1 + x_2)} + 36 = \color{red}{-1} + 6\color{blue}{(-8)} + 36 = -13\end{align}$$
So, you can just finalize it and write the quadratic equation as : 
$$\boxed{\color{brown}{x^2 - 4x - 13}}$$
NOTE : Any quadratic equation can be written as : $x^2 - \color{blue}{S} x + \color{red}{P} $ .
Where: 
   $\color{blue}{S}$ = Sum of Zeroes(or roots) of the equation.
$\color{red}{P}$ = Product of Zeroes(or roots) of the equation.
Hope it helps!
