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

\(x^2 +\dfrac{b}{a}x +\dfrac{c}{a}= (0)\), \(a≠0\) | Dividing both sides of equation by \(a\). |

\(x^2 +\dfrac{b}{a} x=\dfrac{-c}{a}\) | Subtracting \(\dfrac{c}{a}\) from both sides |

\(x^2 +\dfrac{b}{a} x +\bigg(\dfrac{b}{2a}\bigg)^2=\dfrac{-c}{a}+\bigg(\dfrac{b}{2a}\bigg)^2\) | Adding \(\bigg(\dfrac{b}{2a}\bigg)^2\) to both sides; the square of \(\dfrac{1}{2}b\) to get a trinomial square. |

\(\bigg(x \dfrac{b}{2a}\bigg)^2 = \dfrac{-c}{a}+\bigg(\dfrac{b}{2a}\bigg)^2\) | Factoring the left side |

\(\bigg(x+\dfrac{b}{2a}\bigg)^2=\bigg(\dfrac{b}{2a}\bigg)^2+\dfrac{-c}{a}\) | Commutative property of addition on the right side |

\(\bigg(x+\dfrac{b}{2a}\bigg)^2=\bigg(\dfrac{b}{2a}\bigg)^2+ \dfrac{-c}{a}\cdot\dfrac{4a}{4a}\) | To combine the fractions on the right, multiply \(\dfrac{-c}{a}\) by \(4a\) because the LCD is \(4a^2\). |

\(\bigg(x+\dfrac{b}{2a}\bigg)^2= \dfrac{b^2-4ac}{4a^2}\) | Combining the fractions on the right |

\(\sqrt{(x+\dfrac{b}{2a})^2}=\sqrt {\dfrac{b^2-4ac}{4a^2}}\) | Take the square root of both sides |

\(x+\dfrac{b}{2a} = \dfrac{\pm\sqrt{b^2-4ac}}{2a}\) | Square root of both sides |

\(x+\dfrac{b}{2a}-\dfrac{b}{2a} = \dfrac{\pm\sqrt{b^2-4ac}}{2a}-\dfrac{b}{2a}\) | to Solve for \(x\) subtract \(\dfrac{b}{2a}\) from both sides |

\(x =\dfrac{\pm\sqrt{b^2-4ac}}{2a}-\dfrac{b}{2a}\) | Simplifying the left side. |

\(x =-\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}\) | Commutative property on the right side of the equation. |

\(x =-\dfrac{-b±\sqrt{b^2-4ac}}{2a}\) | Combining the fractions |

Return to Watch (Explaining each step)