shit's basic math bruh but your time would be better spent going to khan academy

A quadratic equation is an equation that can be rewritten into the canonical form

0 = a*x^2 + b*x + c

where a, b, and c are specific constants and x is a free variable. A quadratic equation has at most two solutions (a solution is a value of x that makes the equation true) in the real set, but it may have one or no real solutions.

Any quadratic equation in the canonical form can be solved using the quadratic formula:

-b + (+/-square_root(b^2 - 4*a*c) / (2*a))

Note: if b^2 - 4*a*c < 0, then the equation has no real solutions. If it's zero then it has a single real solution. Otherwise it has two real solutions, given by the positive and negative square root.

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>>16490
ax^2+bx+c=0

So you can see everything is on one side and on the other we got 0 so think about this like "where is the function equal to zero [f(x)=0]"
Look at the picture, you can see that there are 3 possibilities; 2 points at which function is equal to 0, 1 point (only the tip of parabola) or there are none. We solve it a bit differently in my country (I think in easier to understand form) so I'll show you how we are thought. First you look at 'a', a can't be equal to 0 (it would be linear equation if a=0), ok 'a' isn't 0 , that means we got a parabola, we can determine whether there are 2, 1 or 0 intersections with the X axis by calculating Δ (capital greek delta)
Δ=b^2 - 4ac
ok so you calculated the Δ, if the Δ<0 that means no intersections with X axis, Δ=0 1 intersection with X axis (the tip) and if Δ>0 we got 2 intersections with the X axis. When you got Δ<0 the equation can't be solved (no points at which function is equal to 0), if Δ=0 (only the tip is touching the X axis) you calculate the X coordinate of the tip, formula for that is (-b)/2a; if Δ>0 we must do a little bit more to solve the equation, because for Δ>0 there are 2 points touching the x axis, we use 2 formulas (-b+√Δ)/(2a) and (-b-√Δ)/(2a) similar to standard quadratic formula.

let's do some examples then

x^2 + 2x + 1 = 0

1) Δ= b^2 - 4ac = 4 - 4*1*1 = 0
2) Δ = 0 so we use formula for the X coordinate of the tip, (-b)/2a = (-2)/(2*1) = (-2)/2 = -1
3) f(x) = 0 if x = 1
4) we can check if it's true, (-1)*(-1) - 1*2 + 1 = 0

10x^2 + x + 100 = 0
1) Δ= b^2 - 4ac = 1 - 4*10*100 = 1 - 4000 = -3999
2) Δ<0, equation can't be solved

3x^2 + 10x + 3 = 0
1) Δ= b^2 - 4ac = 100 - 4*3*3 = 100 - 36 = 64
2) Δ>0
3) (-b-√Δ)/2a = (-10 -8)/6 = (-18)/6 = -3
(-b+√Δ)/2a = (-10 +8)/6 = (-2)/6 = -(1/3)
4) let's check if it's true
3*(-3)*(-3) - 30 + 3 = 27 - 30 + 3 = 0
3*(-1/3)(-1/3) - (10/3) + 3 = 3*(1/9) - (30/9) + (27/9) = (3/9) - (30/9) + (27/9) = 0

Hope it helps

>>16922
sorry, in the 1st example I wrote (f(x)=0 when x=1)
x=-1

Whoa, I just saw this the other day while I was looking for a comment I made in the seventh page.

>>16922
Dude, you're responding to a question from three years ago. I'm sure OP has since learned what a quadratic equation is, or they don't care anymore.

Your method seems a tad too complex for my liking. Why do you think it's better than simply remembering
x = (-b (+/-)sqrt(b^2 - 4*a*c))/(2*a)
? That's a single formula that solves any quadratic equation in a single step.

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>>16924
>>16490
Now it's even farther in the future and the quadratic formula hasn't changed a bit. I hope OP is ok.