Update Quadratic Equations.md

Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Quadratic Equations.md

@@ -121,6 +121,7 @@ $`\therefore`$ The $`x`$-intercepts are at $`\dfrac{2}{3}, \dfrac{-1}{4}`$
 ### Question 3 c)
 
 When $`P(x) = 0`$, that means it is the break-even point for a value of $`x`$ (no profit, no loss).
+
 $`2k^2 + 12k - 10 = 0 \implies k^2 -6k + 5 = 0`$
 
 $`(k-5)(k-1) = 0`$
@@ -158,7 +159,7 @@ $`x = \dfrac{3 \pm \sqrt{23}}{7}`$
 $`\because \dfrac{1}{3}`$ and $`\dfrac{-2}{3}`$ are the roots of a quadratic equation, that must mean that $`(x-\dfrac{1}{3})(x-\dfrac{2}{3})`$ is a quadratic equation that 
 gives those roots.
 
-After expanind we get:
+After expanding we get:
 
 $`y = x^2 - \dfrac{2}{3}x - \dfrac{1}{3}x + \dfrac{2}{3}`$
 
@@ -238,15 +239,10 @@ Since we know that $`(4, 5)`$ is also a point on the parabola, we can susbsitute
 $`5 = a + c \quad (2)`$
 
 ```math
-
 \begin{cases}
-
 9a + c = 0 & \text{(1)} \\
-
 a + c = 5 & \text{(2)} \\
-
 \end{cases}
-
 ```
 
 $`(2) - (1)`$