Solve 18x=45+3x^2
like maybe
3 x^2 -18 x + 45 = 0 ?
or
x^2 - 6 x + 15 = 0
https://www.mathsisfun.com/quadratic-equation-solver.html
3x²+45=18x
Reduce by 3
x²-6x+15=0
Just from here
a=1 b=-6 c=15
Evaluate into the quadratic formula
x=[-b±√(b²-4ac)]/2
I bet you have a typo because I suspect you have not covered complex roots.
Isaac — Please do not use tutors' names in the space for your name. Thanks.
To solve the equation 18x = 45 + 3x^2, we need to rearrange it to bring all the terms to one side of the equation. Let's follow these steps:
1. Start with the given equation: 18x = 45 + 3x^2.
2. Subtract 18x from both sides of the equation to move all the terms to the left side: 18x - 18x = 45 + 3x^2 - 18x.
Simplifying gives us: 0 = 45 - 18x + 3x^2.
3. Rearrange the terms in descending order of powers: 3x^2 - 18x + 45 = 0.
Now, we have a quadratic equation in the form of ax^2 + bx + c = 0, where a = 3, b = -18, and c = 45. To solve this quadratic equation, we can use the quadratic formula, which is:
x = (-b ± √(b^2 - 4ac)) / (2a).
By substituting the values of a, b, and c from our equation into the quadratic formula, we get:
x = (18 ± √((-18)^2 - 4 * 3 * 45)) / (2 * 3).
Simplifying further gives us:
x = (18 ± √(324 - 540)) / 6,
x = (18 ± √(-216)) / 6.
At this point, we notice that the expression under the square root is negative (-216). This means that there are no real solutions to this equation since we can't take the square root of a negative number and obtain a real number.
Therefore, the equation 18x = 45 + 3x^2 has no real solutions.