We know that for quadratic equation \(ax^2+bx+c=0\) , where \(a\) is a non zero number,

we can use the ABC Formula to determine the value of x.

Here is how we get the value of x.

\(ax^2+bx+c=0\)

\(x^2+\frac{b}{a}x+\frac{c}{a}=0\)

\(x^2+\frac{b}{a}x=-\frac{c}{a}\)

\(x^2+\frac{b}{a}x+(\frac{b}{2a})^2=-\frac{c}{a}+(\frac{b}{2a})^2\)

\((x+\frac{b}{2a})^2=-\frac{c}{a}+(\frac{b}{2a})^2\)

\((x+\frac{b}{2a})^2=-\frac{c}{a}+\frac{b^2}{4a^2}\)

\((x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\)

\(x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\)

\(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

Then how about cubic equation \(ax^3+bx^2+cx+d=0\) ?

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TopNewestYou will get your answers here – Sambhrant Sachan · 1 year, 2 months ago

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– Andy Leonardo · 1 year, 2 months ago

Thankyou Sambhrant , it's really helpful.Log in to reply