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Mitiko63558yThere are 2 solutions:
x^2 + 2x -3 = 0;
x^2 - x + 3x - 3 = 0;
x(x-1) + 3(x-1) = 0;
(x + 3)(x - 1) = 0;
X = 1 || X = -3; -
Mitiko63558y@zero1 Every square equation has two answers, so he got half of them. One more to go...
If he does this one, I'm gonna prove him as math god. -
@LicensedCrime I find the binomial way easier for this kind of equation, it is solve almost by seeing it
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@LicensedCrime Exactly, the key was "kind" what was sad is that it took me a while to remember š
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Really fucking bad at math and don't understand this at all but the bleeding eyes picture is funny enough and gives some context š
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Roots =(-b +,- sqrt(b*b-4bc))/2a
So I quad equation ax²+bx+c has two solutions.
It maybe real or imaginary....
Peaceā -
@LicensedCrime @Bizzarist
For an quadratic equation of type ax²+bx+c=0, the determinant D is given as
D=b²-4ac;
If D is +ve,then both roots are distinct and real.
If D is 0 , then both roots are equal and will be real....
If D is -ve ,then both roots will be imaginary (real numbers can also be represented as imaginary numbers)
If we want to find the actual roots then we can use the equation I mentioned in the previous post...
Peaceā -
@thmnmlst I went and checked out all your rants only to see what @jpichardo was talking about š
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iAmNaN71318yOh my. I don't think that is how it's supposed to work. I got to line 2 and was like, "waaaaa?"
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That person who wrote the bs logic is either 1)lucky to have solved for 1 solution
2)a person who actually knew how to do it and knew the final answers and wanted to screw around and write random stuff and manipulate it to one of the correct solutions
When you write BS logic, but the code works anyways...
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