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Hi friends this is a small tutorial regarding the proof of quadratic equation ... this is made by me

Mathematics
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Quadratic Formula*?
yes @terenzreignz \[\Huge{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}\]

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Other answers:

\[{\textbf{the general form of a quadratic equation is ax^2+bx+c=0 . .}}\] \[\textbf{it is known as quadratic equation since it's degree is 2 }\] HOW TO GET THE ROOT OF A QUADRATIC EQUATION : \[\huge{ax^2+bx+c=0}\] \[\huge{ax^2+bx=-c}\] \[\huge{\frac{(ax^2+bx)}{a}=\frac{-c}{a}}\] \[\huge{x^2+\frac{bx}{a}=\frac{-c}{a}}\] \[\huge{(x)^2+2(\frac{b}{2a})(x)=\frac{-c}{a}}\] \[\huge{(x)^2+2(\frac{b}{2a})(x)+\frac{b^2}{4a^2}=\frac{-c}{a}+\frac{b^2}{4a^2}}\] \[\huge{(x+\frac{b}{2a})^2 = \frac{-c}{a}+\frac{b^2}{4a^2}}\] \[\huge{x+\frac{b}{2a}=\pm\sqrt{\frac{-c}{a}+\frac{b^2}{4a^2}}}\] \[\huge{x=\pm\sqrt{\frac{-c}{a}+\frac{b^2}{4a^2}}-\frac{b}{2a}}\] \[\huge{x=\pm\sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}}\] \[\huge{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}\]
Good work, mathslover, I appreciate your tutorial :)
thanks a lot @ParthKohli
Please prefer the LaTeX for better proof and for no mistakes
Great job @mathslover!
You are very welcome
@mathslover wat is ur intention by posting this!
to just make you all aware with quadratic equation
ok
I am going to post some more interesting things about this
plz go ahead
i will but after some time. .. doing some research work :)
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@wdkekefkjrgjr dont post such comments
i m posting more interesting things about quadratic equation as a new question
k! thanx a lot @mathslover :)
Ur welcome @jiteshmeghwal9 Best of Luck

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