Question

come up with an equation that would relate to a practical situation

Answer 1

I will try and post again, for some reason this web page keeps jumping around.

Answer 2

Well that posted. I was going to develop an equation that will solve quadratic equations using the terms coefficients and the constant. The general form of a quadratic is:
Ax^2 + Bx + C = 0 Where A is coefficient of the leading term whose variable is to the 2nd power, B is the coefficient of x, the variable whose exponent is 1 or is normally not show, C is the constant. The equation will be called the "Quadratic Formula" This has been done years ago but just to show you how it is a practical use for an equation.

Answer 3

Completing the Square" is one way to solve a quadratic. Using that method will lead to the equation that we are looking for: Ax^2 +Bx + C = 0 is our quadratic equation.
Step 1. Is to divide the x^2 coefficient by itself (A) so that its coefficient a "1" of course to keep the equation balanced all the remaining terms will also be divided by A.

You now have: X^2 + (B/A)x +C/A = 0
Step 2. Subtract the C/A term from both sides getting: x^2 + (B/A)x = -C/A Step 3. Just as in "completing the square" procedure, take 1/2 of the coefficient of the x term (now(B/A), or after halving it (B/(2A)) then squaring it and adding it to both sides the equation.
Getting x^2 + (B/A)x + B^2/(4A^2) = B^2/(4A^2) -C/A

Step 4. On the left of the equal sign we now have a perfect square. Express the equation showing that relationship (x + B/(2A))^2 = B^2/(4A^2) - C/A

Step 5. Do the operation on the right of the equal sign. Combine those two fractions, the LCD is 4A^2 you now have: [x + B/(2A]^2 = (B^2-4AC)/4A^2

Step 6. Take the square root of both sides.. We now have x + B/(2A)=[sqrtB^2-4AC]/(2A)
Step 7. Subtract the B/(2A) term from both sides. You now have:
x=[-B +/- sqrt(B^2-4AC)]/2A
There you have it, the "Quadratic Formula" an equation with practical application.