Question

Solve for s.
s2+12s–13=0
Write each solution as an integer, proper fraction, or improper fraction in simplest form. If there are multiple solutions, separate them with commas.

Answer 1

AAAaannndd the bot gets it wrong yet again!
how can you flub such easy calculation?
s = (-12 - 14) / 2 = -26/2 = -13

or, with less work,
s^2+12s–13=0
(s+13)(s-1) = 0
s = -13, 1

Answer 2

Using the quadratic formula:

s = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 1, b = 12, and c = -13. Plugging in:

s = (-12 ± √(12^2 - 4(1)(-13))) / 2(1)
s = (-12 ± √(144 + 52)) / 2
s = (-12 ± √196) / 2
s = (-12 ± 14) / 2

So the solutions are:

s = (-12 + 14) / 2 = 1
s = (-12 - 14) / 2 = -13/2

Therefore, the solutions are:

s = 1, -13/2

Answer 3

I apologize for the error in my previous response. You are correct that the solutions are s = -13 and s = 1. Using the formula:

s = (-b ± √(b^2 - 4ac)) / 2a

we get:

s = (-12 ± √(12^2 - 4(1)(-13))) / 2(1)
s = (-12 ± √(144 + 52)) / 2
s = (-12 ± √196) / 2
s = (-12 ± 14) / 2

which simplifies to:

s = (-12 + 14) / 2 = 1
s = (-12 - 14) / 2 = -13

Thank you for bringing this to my attention and I will make sure to double-check my calculations in the future.