Obviously I don't have any mathematical or scientific training or whatever so this could very well belong to a sqt, but isn't basic algebra (probably not a proper math object) supposed to be transitive? Like if a=b and a=c then b=c? Then why is (-b+(b^2-4ac)^(1/2))/2a) ≠ (-b-(b^2-4ac)^(1/2))/2a) even though they both are equal to ax^2+bx+c=0)? Also there's the if a=b then a+-*/c=b+-*/c property. It doesn't hold for "x=7 then x/(x-7)=7(x-7)" Does this mean math is broken? Do we just patch it up because it's useful irl? Doesn't the fact that there are undefined operations in basic maths prove that it is broken? Wasn't something like this, an undefined result, what killed, for instance, type theory?
>>8101179
You can't equate an expression with an equation.
>>8101179
If we take b=0, a = -1, and c = some positive number, then your question translates to
"if (-1)*x^2 + (0)*x + c = 0, and if x = ((-(0)+((0)^2-4(-1)c)^(1/2))/2(-1)) or ((-(0) - ((0)^2-4(-1)c)^(1/2))/2(-1)) , then shouldn't ((-(0)+((0)^2-4(-1)c)^(1/2))/2(-1)) = ((-(0) - ((0)^2-4(-1)c)^(1/2))/2(-1)) be true? No, if x is this number, OR that number, it does not mean it is this number AND that number.
-2^2=4 and 2^2=4 doesn't mean -2=2, so no cigar.
>>8101179
I really do not get this picture's popularity at all.
There's a cigar in the wall. So what.
>>8101202
Besides 1/3(.333333) to 3/3(1.0)
the quadratic equation is solving for x values that make the ax^2 + bx + c expression equal to 0
x = (-b +/- (b^2-4ac))/2a
x =/= ax^2 +bx + c
>>8101179
They are not literally equal to that quadratic. Why do you think that? That doesn't make sense in the least. If you sub $(-b + \sqrt{b^{2} -4ac})/2a$ for x into $ax^{2} + bx + c$ you get zero. That's what that means.
>>8101208
this is the reason op
Hahaha wtf, did you skip highschool?
