I have an odd problem with no solution. I am completely lost on how to solve this.

Problem:

Find the coordinates of the point(s) of intersection of the line $\displaystyle x = 1+t, y = 2+3t, z = 1-t$

and the surface $\displaystyle z = x^2 +2y^2$

Attempt:

(1) $\displaystyle x = 1+t$

(2) $\displaystyle y = 2+3t$

(3) $\displaystyle z = 1-t$

Subbing in (1), (2), (3) into the surface i have

$\displaystyle 1-t = (1+t)^2 +2(2+3t)^2$

Solving i get:

$\displaystyle 19t^2+27t+8=0$

At this point i need to solve for t

It appears to be a quadratic equation.

$\displaystyle \begin{array}{*{20}c} {t = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}$

$\displaystyle \begin{array}{*{20}c} {t = \frac{{ - 27 \pm \sqrt {(27)^2 - 4(19)(8)} }}{{2(19)}}} \end{array}$

$\displaystyle \begin{array}{*{20}c} {t = \frac{{ - 27 \pm \sqrt {121} }}{{38}}} \end{array}$

$\displaystyle \frac{{ - 27 \pm 11 }}{{38}} $

$\displaystyle \frac{{ - 27 + 11 }}{{38}} $ and $\displaystyle \frac{{ - 27 - 11 }}{{38}} $

$\displaystyle t= \frac{{ - 8 }}{{19}} $ and $\displaystyle t = -1 $

Then just plug these t values into the parametric equations to get the points?

Did i do this right?