How do you find quadratic equations if you are only gven the solution?

suppose you know the solutions are 5 and -3

form factors (x-5) and (x+3), multiply them and set the result equal to zero

(x-5)(x+3)=0
x^2 -2x -15=0

Is it possible to have diferent quadratic equations with the same solution? My instinct says yes, my fear of math says no.

no

any quadratic equation, which has been simplied to its simplest form has a unique solution, and given an set of solutions there is only one unique quadratic equation

If I take the equation from my example and multiply it by 3 I would get
3x^2 - 6x - 45 = 0

it really is not a "different" equation.

BTW, if your solutions had been fractions, say 4/5 and -2/3, your two factors would have been
(5x-4) and (3x+2)

There is another way:
If you know the two solutions are m and n, then you can just form the quadratic by writing

x^2 - (m+n)x + mn = 0, then simplify

eg. going back to my example of solutions of 5 and -3, their sum is 2 and their product is -15

the equation would be x^2 -(2)x + (15) = 0 , just like above

change

"the equation would be x^2 -(2)x + (15) = 0 , just like above" to

the equation would be x^2 -(2)x + (-15) = 0 , just like above

Wow! I have a lot to learn. I don't mean to take up your time, but I understand there is more than one way to solve quadratic equations. Is there a way you think is easier and how?

To find a quadratic equation if you are only given the solution, you need to know the roots or the values of x for which the equation equals zero. Let's say the given solutions are x = a and x = b.

To find the quadratic equation, you can use the fact that the roots of a quadratic equation are related to its coefficients. The general form of a quadratic equation is:

ax^2 + bx + c = 0

Where a, b, and c are coefficients that need to be determined.

Given the solutions x = a and x = b, you can use Vieta's formulas. Vieta's formulas state that for a quadratic equation:

a + b = - (coefficient of x) / (coefficient of x^2)
ab = constant term / (coefficient of x^2)

By substituting the given values into these formulas, you can derive the values of a, b, and c, and thus obtain the quadratic equation.