Solve, Part 2

Thanks to the Quadratic Formula (that I always seem to forget about...) the problem I had trouble with can now be solved. And could have been solved in last semester's class. 95% of the time I forget about this never-fail formula.

Since I can't write it out as it should appear, you'll have to discern it is written.

The Quadratic Formula:

(-b +/- *the square root of* b^2 - 4(a)(c)) / 2(a)

The (so far) answer we had then was -3(a^2+2a+11)=0

First, you'll take the -3 and simplify it: -3=0. Um, nope! So that eliminates that.

Then, you'll take the remaining: a^2+2a+11. To fill in the polynomial in the formula we know that a=1, b=2 and c=11. Use those numbers and fill in the blanks.

(-2 +/- *the square root of* 2^2-4(1)(11)) / 2(1)

That simplifies into, (try to keep up): (-2 +/- *the square root of* -40) / 2

Simplify the square root of -40 and you'll get: -2 +/- 2i *the square root of* 10 / 2

And all that will simplify down to: -1 +/- i *square root of* 10

Any mathematicians need to correct me? Please feel free to do so, because I'm still not 100% positive this is correct.