Question

I am not very good at this; can someone help me out?

Solve the equation t² + 12t + 13 = 0. The answer should look something like 6 + √23, and there should be two of them.
A) 6 + √23 and 6 - √23
B) 3 + √5 and 3 - √5
C) 2 + √10 and 2 - √10
D) 7 + √20 and 7 - √20

Answer 1

To solve the equation t² + 12t + 13 = 0, we can use the quadratic formula. The solutions are t = -6 + √23 and t = -6 - √23.

Step-by-step explanation:

To solve the equation t² + 12t + 13 = 0, we can use the quadratic formula. The quadratic formula is given by:

t = (-b ± √(b² - 4ac)) / (2a)

Substituting the values a = 1, b = 12, and c = 13 into the formula, we have:

t = (-12 ± √(12² - 4(1)(13))) / (2(1))

t = (-12 ± √(144 - 52)) / 2

t = (-12 ± √92) / 2

Simplifying further, we have:

t = (-12 ± 2√23) / 2

t = -6 ± √23

Therefore, the solutions to the equation are t = -6 + √23 and t = -6 - √23.