It is often useful to know the roots of a quadratic function. We will first derive an expression for the inverse of a quadratic polynomial $$y(x)$$ and then find the roots by setting $$y=0$$. The general form of a quadratic polynomial is $$! \displaystyle y = ax^2 + bx + c$$ We wish to solve for $$x$$. The most straightforward way to accomplish this is by completing the square. First we divide the equation by the coefficient of the quadratic term, $$a$$, $$! \displaystyle \frac{y}{a} = x^2 + \frac{b}{a}x + \frac{c}{a}$$ Now move all of the terms involving $$x$$ to the left side and move everything else to the right side of the equation. $$! \displaystyle x^2 + \frac{b}{a}x = \frac{y}{a} – \frac{c}{a}$$ The left hand side can be made into a perfect square by adding the square of half of the coefficient of the linear term $$! \displaystyle x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2= \left(\frac{b}{2a}\right)^2 + \frac{y}{a} – \frac{c}{a}$$ then the left can be written as $$! \displaystyle \left(x + \frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 + \frac{y}{a} – \frac{c}{a}$$ Now re-write the right side by finding a common denominator, $$! \displaystyle \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 + 4ay – 4ac}{4a^2}$$ $$! \displaystyle \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 + 4a(y – c)}{4a^2}$$ taking the square root of both sides, $$! \displaystyle x + \frac{b}{2a} = \frac{\pm\sqrt{b^2 + 4a(y – c)}}{2a}$$ Now isolate $$x$$ to get the inverse of a quadratic $$! \displaystyle x = \frac{-b\pm\sqrt{b^2 + 4a(y – c)}}{2a}$$ The special case when $$y=0$$ yields the quadratic formula $$! \displaystyle x = \frac{-b\pm\sqrt{b^2 – 4ac}}{2a}$$