Question

find the numbers such that the square of number is 15 lesser than twice the number.​

Answer 1

Answer:

no real answer

Step-by-step explanation:

algebra:

x^2 + 15 = 2x

x^2 - 2x + 15 = 0

quadratic formula:

x = (-b + sqrt(b^2 - 4ac)) / 2a   and   x = (-b - sqrt(b^2 - 4ac)) / 2a

x = (2 + sqrt(-56)) / 2   and   x = (2 - sqrt(-56))) / 2

*notice sqrt(-56) requires complex numbers

complex answer:

x = 1 + sqrt(14)i   and   x = 1 - sqrt(14)i