Question

what are quadratics, and a quadratic equation? How would you solve one?

Answer 1

They are equations of the type
ax^2 + bx + c = 0, where a, b, and c are real numbers,
with a ≠ 0 , or else the ax^2 wouldn't be there.

e.g. 2x^2 + 12x - 7 = 0

"Solving them" consists of finding where the corresponding parabola intersects the x-axis

The solution could consist of 3 cases:
- the equation has 2 different real solutions, when the parabola cuts the
x-axis at two different places
- the equation has 1 real solution, when the parabola touches the x-axis
- the equation no real solutions, when the parabola does not intersect
the x-axis. In that case the equation has 2 distinct imaginary or complex
solutions of the form p ± q i, where i = √-1

There are 3 main methods to solve:
1. Using the quadratic formula, x = (-b ± √(b^2 - 4ac) )/(2a)
this method always works and becomes the default methods for
many students
2. Factoring, sometimes works
3. Completing the square, always works

e.g
8x^2 + 2x - 15 = 0

- by formula, here a = 8, b = 2, and c = -15
x = (-2 ±√(4 - 4(8)(-15) )/16
x = (-2 ±√484)/16
= (-2 ± 22)/16
x = 20/16 or x = -24/16
x = 5/4 or x = -3/2

If you get rational roots, like I just did, the equation will factor
(If one root is rational, the other one will also be rational. If one is
irrational, the other one will be as well.

- factoring (will not work if roots are irrational, but you won't know that
until you try it)
8x^2 + 2x - 15 = 0
(2x+3)(4x-5) = 0
x = -3/2 or x = 5/4, just like we did before

- completing the square:
8x^2 + 2x - 15 = 0
divide each term by 8, to get x^2 at the front, moving the constant to the
right side of the equation
x^2 + (1/4)x = 15/8
x^2 + (1/4)x + 1/64 = 15/8 + 1/64, I assume you know how this works
(x + 1/8)^2 = 121/64
x+1/8 = ± 11/8
x = -1/8 ± 11/8 = 5/4 or -3/2 , as before

I suggest you print my reply to study it carefully, using some of your
own examples

Answer 2

Honestly, thank you so much.