Question

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.

Answer 1

The answer is correct if it was meant to be, after cancelling the two's:
5±√(10)

Answer 2

The answer is correct if it was meant to be (without the 2 in the denominator):
5±2√(10)

Answer 3

Where does that two in the denominator go? Since it has to cancel out with all terms.

Answer 4

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.