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

Solutions: $$ax^2+bx+ c=0$$

 $$x^2 +\Large\frac{b}{a}\normalsize x +\Large\frac{c}{a}= \normalsize(0)$$, $$a≠0$$ Dividing both sides of equation by a. $$x^2 +\Large\frac{b}{a}\normalsize x=\Large\frac{-c}{a}$$ Subtracting $$\frac{c}{a}$$ from both sides $$x^2 +\Large\frac{b}{a}\normalsize x +(\Large\frac{b}{2a})^2=\Large\frac{-c}{a}+(\Large\frac{b}{2a})^2$$ Adding $$(\Large\frac{b}{2a})^2$$ to both sides; the square of $$\frac{1}{2}b$$ to get a trinomial square. $$(x + \Large\frac{b}{2a})^2 = \Large\frac{-c}{a}+(\Large\frac{b}{2a})^2$$ Factoring the left side $$(x + \Large\frac{b}{2a})^2=(\Large\frac{b}{2a})^2+\Large\frac{-c}{a}$$ Commutative property of addition on the right side $$(x + \Large\frac{b}{2a})^2=(\Large\frac{b}{2a})^2+ \Large\frac{-c}{a}•\Large\frac{4a}{4a}$$ To combine the fractions on the right, multiply $$\Large\frac{-c}{a}$$ by $$4a$$ because the LCD is $$4a^2$$. $$(x + \Large\frac{b}{2a})^2= \Large\frac{b^2-4ac}{4a^2}$$ Combining the fractions on the right $$\sqrt{(x + \Large\frac{b}{2a})^2}=\sqrt {\Large\frac{b^2-4ac}{4a^2}}$$ Take the square root of both sides $$x + \Large\frac{b}{2a} = \Large\frac{±\sqrt{b^2-4ac}}{2a}$$ Square root of both sides $$x + \Large\frac{b}{2a}-\Large\frac{b}{2a} = \Large\frac{±\sqrt{b^2-4ac}}{2a}-\Large\frac{b}{2a}$$ to Solve for $$x$$ subtract $$\Large\frac{b}{2a}$$ from both sides $$x =\Large\frac{±\sqrt{b^2-4ac}}{2a}-\Large\frac{b}{2a}$$ Simplifying the left side. $$x =-\Large\frac{b}{2a}±\Large\frac{\sqrt{b^2-4ac}}{2a}$$ Commutative property on the right side of the equation. $$x =-\Large\frac{-b±\sqrt{b^2-4ac}}{2a}$$ Combining the fractions