Solve for x=
x^2 + 45= 6x
x^2 - 6x = -45
x^2 - 6x + 36 = -45 + 36 , I completed the square
(x-6)^2 = -9
x-6 = ±√-9
x = 6 ± 3i
Thanks for the help!
To solve for x in the equation x^2 + 45 = 6x, we need to rearrange the equation so that the terms involving x are on one side and the constant term is on the other side.
First, subtract 6x from both sides of the equation:
x^2 + 45 - 6x = 0
This gives us a quadratic equation in the standard form:
x^2 - 6x + 45 = 0
Now, there are multiple ways to solve a quadratic equation. One common method is to use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, the equation is in the form ax^2 + bx + c = 0, so we can identify a = 1, b = -6, and c = 45.
Plugging these values into the quadratic formula, we get:
x = (-(-6) ± √((-6)^2 - 4*1*45)) / (2*1)
Simplifying further, we have:
x = (6 ± √(36 - 180)) / 2
x = (6 ± √(-144)) / 2
At this point, we encounter a problem. The expression √(-144) involves the square root of a negative number, which is not real. Therefore, there are no real solutions to the equation x^2 + 45 = 6x.