# POLQ 2 | Lesson 4 | Try This! (Deriving the Quadratic Formula)

The standard form of the quadratic equation is $$ax^2+bx+c=0$$. Use your understanding of completing the square to put the steps below in the correct order to derive the quadratic formula.
 $$\left( x + \dfrac{b}{2a} \right) = \pm \sqrt{ \dfrac{b^2-4ac}{4a^2}}$$ $$x^2 + \dfrac{b}{a} x + \bigg( \dfrac{b}{2a} \bigg)^2 = – \dfrac{c}{a}+\bigg( \dfrac{b}{2a}\bigg)^2$$ $$\sqrt{\bigg( x + \dfrac{b}{2a}\bigg)^2} = \pm \sqrt{ \dfrac{b^2-4ac}{4a^2}}$$ $$x = \dfrac {-b \pm \sqrt{b^2-4ac} } {2a}$$ $$\bigg( x + \dfrac{b}{2a} \bigg)^2 = -\dfrac{c}{a} \cdot \dfrac{4a}{4a} + \dfrac{b^2}{4a^2}$$ $$x + \dfrac{b}{2a} = \pm \dfrac{\sqrt{ b^2-4ac}}{2a}$$ $$x^2 + \dfrac{b}{a} x + \dfrac{c}{a} = 0$$,  $$a \neq 0$$ $$x = -\dfrac{b}{2a} \pm \dfrac{\sqrt{b^2-4ac}}{2a}$$ $$\bigg( x + \dfrac{b}{2a}\bigg)^2 = -\dfrac{c}{a}+\bigg( \dfrac{b}{2a}\bigg)^2$$ $$x + \dfrac{b}{2a} = \pm \dfrac{\sqrt{b^2-4ac}}{ \sqrt{4a^2}}$$ $$\bigg( x + \dfrac{b}{2a}\bigg)^2 = -\dfrac{c}{a} + \dfrac{b^2}{4a^2}$$ $$x^2 + \dfrac{b}{a} x = \ – \dfrac{c}{a}$$ $$\bigg( x + \dfrac{b}{2a}\bigg)^2 = \dfrac{b^2-4ac}{4a^2}$$