Posted by **-Untamed-** on Monday, December 5, 2011 at 7:42pm.

Solve the following quadratic equation(as exact values) using the quadratic formula.

x^2-10x-15=0

I got 5+-2sqrt10/2.

^ Sorry hard to write it out. But basically there is a 5 there, the negative sign is supposed to be under the positive sign, then there is a two outside of the radicand, and inside the radicand is 10, all under 2.

- Math -
**MathMate**, Monday, December 5, 2011 at 8:06pm
The answer is correct if it was meant to be, after cancelling the two's:

5±√(10)

- Math - correction -
**MathMate**, Monday, December 5, 2011 at 8:10pm
The answer is correct if it was meant to be (without the 2 in the denominator):

5±2√(10)

- Math -
**-Untamed-**, Monday, December 5, 2011 at 8:15pm
Where does that two in the denominator go? Since it has to cancel out with all terms.

- Math -
**MathMate**, Monday, December 5, 2011 at 8:52pm
x=(-b±sqrt(b²-4ac))/2a

a=1

b=-10

c=-15

so substituting,

x=(10±sqrt((-10)²-4(1)(-15)))/2

=(10±sqrt(160))/2

=(10±4sqrt(10))/2

=5±2sqrt(10)

You may have (like I did) subtracted 4ac (and got 40) instead of adding to get 160.

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