Quadratic formula to solve equation

I am in the beginning half of college algebra and have one question I am having a hard time finding a solution to:

Use the quadratic formula to solve: 3x^2 - 5x = -8

As far as I get is (V symbolizing the squareroot):
3x^2 - 5x + 8 = 0
x = (5 + -V-5^2 - (4*3*8)) / 2*3
x = (5 + -V25 - 96)/6
x = (5 + -V-71)/6

This is as far as I can get but can't find any real solution. Any help would be appreciated.

Note: For "square root of", please use "sqrt()", with the grouping symbols around whatever is inside the radical. In this case, I have a pretty good idea what you meant, because I know the context, but that won't always be the case.

You have:

. . . . .3x² - 5x = -8
. . . . .3x² - 5x + 8 = 0

. . . . .x = (5 ± sqrt[(-5)² - 4(3)(8)]) / (2×3)
. . . . .x = (5 ± sqrt[25 - 96]) / 6

Since 25 - 96 is negative, there is no "real" solution. If you haven't studied complex numbers yet, then the answer is "no solution"; if you have, then continue on to get the solution containing the "i".

Eliz.

Crookshanks said:

3x^2 - 5x + 8 = 0

Click to expand...

Good.

x = (5 + -V-5^2 - (4*3*8)) / 2*3

Click to expand...

Notation Problem:
-5^2 = -(5^2) = -25
(-5)^2 = 25, which is what you intended.
Also, V(-5)^2 - (4*3*8) may be more clear as V[(-5)^2 - (4*3*8)]

x = (5 + -V25 - 96)/6
x = (5 + -V-71)/6

Click to expand...

25 - 96 = -71
What can you do about that? If you have a good reason to BELIEVE there are Real solutions, check your problem statement. If you find no error, document your findings and move on. Maybe something like this, "I expected Real solutions, but clearly there are none, based on the Negative value under the square root."

"Use the quadratic formula to solve: 3x^2 - 5x = -8"

I suggest you VERIFY that equation; I suspect it should be: 3x^2 - 5x = 8;
then, you'll end up with a "nice" solution...