# POLQ 2 | Lesson 4 | Watch (Explaining each Step) Solutions

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