diff --git a/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Quadratic Equations.md b/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Quadratic Equations.md
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--- a/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Quadratic Equations.md	
+++ b/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Quadratic Equations.md	
@@ -1,4 +1,272 @@
 ### Question 1 a)
 
-The zeroes of a quadratic equation are the solutiosn of $`x`$ when $`ax^2+bx+c = 0`$. The roots of the quadratic equation is when $`(x + r_1)(x + r_2) = 0`$, more 
-commonly described by the formula $`\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}`$. Therefore the roots of a quadratic equation are also the zeroes of the quadratic equation.
\ No newline at end of file
+The zeroes of a quadratic equation are the solutions of $`x`$ when $`ax^2+bx+c = 0`$. The roots of the quadratic equation is when $`(x + r_1)(x + r_2) = 0`$, more 
+commonly described by the formula $`\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}`$. Therefore the roots of a quadratic equation are also the zeroes of the quadratic equation.
+
+### Question 1 b)
+
+$`D = b^2 - 4ac`$
+
+If $`D=0`$, there is one zero.
+
+$`\therefore n^2 - 4(1)(4) = 0`$
+
+$`n^2 - 16 = 0`$
+
+$`(n+4)(n-4) = 0`$
+
+$`n = \pm 4`$
+
+### Question 1 c)
+
+Let $`h`$ be the height and $`b`$ be the base. $`h = 2b + 4`$.
+
+$`\therefore (2b+4)(b) = 168(2)`$
+
+$`4b^2 + 8b = 168(2)`$
+
+$`b^2 + 2b = 84 = 0`$
+
+$`b = \dfrac{-2 \pm \sqrt{4+4(84)}`$
+
+$`b = \dfrac{-2 \pmm \sqrt{340}}{2}`$
+
+$`b = -1 + \sqrt{85}`$
+
+### Question 2 a)
+
+If the quadratic is in $`ax^2 + bx + c`$, the AOS (axis of symmetry) is at $`\dfrac{-b}{2a}`$. And you can plug that value into the quadratic equation to get your optimal value,
+which is:
+
+$`= a(\dfrac{-b}{2a})^2 + b(\dfrac{-b}{2a}) + c`$
+
+$`= \dfrac{b^2}{4a} + \dfrac{-b^2}{2a} + c`$
+
+$`= \dfrac{-b^2}{4a} + c`$
+
+### Question 2 b)
+
+$`2x^2 + 5x - 1 = 0`$
+
+$`2(x^2 + \dfrac{5}{2}x + (\dfrac{5}{4})^2 - (\dfrac{5}{4})^2) - 1 = 0`$
+
+$`2(x+\dfrac{5}{4})^2 - \dfrac{25}{8} - 1 = 0`$
+
+$`2(x+ \dfrac{5}{4})^2 - \dfrac{33}{8} = 0`$
+
+$`(x+ \dfrac{5}{4})^2 = \dfrac{33}{16}`$
+
+$`x = \pm \sqrt{\dfrac{33}{16}} - \dfrac{5}{4}`$
+
+$`x = \dfrac{\pm \sqrt{33}}{4} - \dfrac{5}{4}`$
+
+$`x = \dfrac{\sqrt{33}-5}{4}`$ or $`\dfrac{-\sqrt{33} - 5}{4}`$
+
+### Question 2 c)
+
+Let $`w`$ be the width between the path and flowerbed, $`x`$ be the length of the whole rectangle and $`y`$ be the whole rectangle (flowerbed + path).
+
+$`x = 9+2w`$
+
+$`y = 6+2w`$
+
+$`(6+2w)(9+2w) - (6)(9 = (6)(9)`$
+
+$`54 + 12w + 18w + 4w^2 = 2(54)`$
+
+$`4w^2 + 30w = 54`$
+
+$`2w^2 + 15w - 27 = 0`$
+
+$`2 \quad \quad -3`$ 
+
+$`1 \quad \quad 9`$
+
+$`(2w-3)(w+9) = 0`$
+
+$`w = \dfrac{3}{2}, -9`$
+
+$`\because w \gt 0`$
+
+$`\therefore w = \dfrac{3}{2}`$
+
+$`\therefore x = 9+3 = 12`$
+
+$`\therefore y = 6+3 = 9`$
+
+$`P = 2(x + y) \implies P = 2(12+9) \implies P = 42`$
+
+$`\therefore`$ The perimeter is $`42m`$
+
+### Question 3 a)
+
+Use discriminant, where $`D = b^2 - 4ac`$.
+
+```math
+\begin{cases}
+
+\text{If } D \gt 0 & \text{Then there are 2 real distinct solutions} \\
+
+\text{If } D = 0 & \text{Then there is 1 real solution} \\
+
+\text{If } D \lt 0 &\text{Then there are no real solutions} \\
+
+\end{cases}
+```
+
+### Question 3 b)
+
+$`y = 12x^2 - 5x - 2`$
+
+$`3 \quad \quad -2`$
+
+$`4 \quad \quad 1`$
+
+$`y = (3x-2)(4x+1)`$
+
+$`\therefore`$ 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).
+$`2k^2 + 12k - 10 = 0 \implies k^2 -6k + 5 = 0`$
+
+$`(k-5)(k-1) = 0`$
+
+$`k = 5, 1`$
+
+Either $`5000`$ or $`1000`$ rings must be produced so that there is no prodift and no less.
+
+AOS (axis of symmetry)  = $`\dfrac{-b}{2a} = \dfrac{6}{2} = 3`$ 
+
+$`\therefore 3000`$ rings should be made to achieve the optimal value.
+
+Maximum profit $`= -2(3)^2 + 12(3) - 10`$
+
+$`= -18 + 30 - 10`$
+
+$` = 8`$
+
+$`\therefore 8000`$ dollars is the maximum profit.
+
+### Question 4 a)
+
+$`5x(x-1) + 5 = 7 + x(1-2x)`$
+
+$`5x^2 - 5x = 2 + x - 2x^2`$
+
+$`7x^2 - 6x - 2 = 0`$
+
+$`x = \dfrac{6 \pm \sqrt{92}}{14}`$
+
+$`x = \dfrac{3 \pm \sqrt{23}}{7}`$
+
+### Question 4 b).
+
+$`\because \dfrac{1}{3}`$ 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:
+
+$`y = x^2 - \dfrac{2}{3}x - \dfrac{1}{3}x + \dfrac{2}{3}`$
+
+$`y = x^2 - x + \dfrac{2}{3}`$
+
+Now we complete the square.
+
+$`y = x^2 - x + (\dfrac{1}{2})^2 - (\dfrac{1}{2})^2 + \dfrac{2}{3}`$
+
+$`y = (x-\dfrac{1}{2})^2 - \dfrac{1}{4} + \dfrac{2}{3}`$
+
+$`y = (x-\dfrac{1}{2})^2 + \dfrac{5}{12}`$
+
+### Qustion 4 c)
+
+When $`h = 0`$, the ball hits ground, so:
+
+$`-3.2t^2 + 12.8 + 1 = 0`$
+
+$`t = \dfrac{-12.8 \pm \sqrt{12.8^2 - 4(3.2)}}{-6.4}`$
+
+$`t = \dfrac{12.8 \pm \sqrt{151.04}}{6.4}`$
+
+$`\because t \ge 0`$
+
+$`\therefore t = \dfrac{12.8 \pm sqrt{151.04}}{6.4}`$
+
+$`\therefore t \approx 3.9`$
+
+The ball will strike the ground at approximately $`3.9`$ seconds.
+
+### Question 5 a)
+
+$`128 = 96t - 16t^2`$
+
+$`16t^2 - 96t + 128 = 0`$
+
+$`t^2 - 6t + 8 = 0`$
+
+$`1 \quad \quad -2`$
+
+$`1 \quad \quad -4`$
+
+
+$`(t-2)(t-4)`$, at seconds $`2`$ and $`4`$, the rocket reaches $`128m`$.
+
+### Question 5 b)
+
+Break even is when revenue = cost.
+
+$`\therefore R(d) = C(d)`$
+
+$`-40d^2 + 200d = 300 - 40d`$
+
+$`40d^2 - 240d + 300 = 0`$
+
+$`4d^2 - 24d + 30 = 0`$
+
+$`2d^2 - 12d + 15 = 0`$
+
+$`d = \dfrac{12 \pm \sqrt{24}}{4}`$
+
+$`d = \dfrac{6 \pm \sqrt{6}}{2}`$
+
+At $`d = 4.22474407$ or $`1.775255135`$ is when you break even.
+
+### Question 5 c)
+
+Let the quation be $`y = a(x-d)^2 + c`$
+
+Since we know that that $`(0, 0)`$ and $`(6, 0)`$ are the roots of this equation, the AOS is when $`x = 3`$
+
+$`\therefore y = a(x-3)^2 + c`$.
+
+Since we know that $`(0, 0)`$ is a point on the parabola, we can susbsitute it into our equation.
+
+$`0 = 9a + c \quad (1)`$
+
+Since we know that $`(4, 5)`$ is also a point on the parabola, we can susbsitute it int our equation as well.
+
+$`5 = a + c \quad (2)`$
+
+```math
+
+\begin{cases}
+
+9a + c = 0 & \text{(1)} \\
+
+a + c = 5 & \text{(2)} \\
+
+\end{cases}
+
+```
+
+$`(2) - (1)`$
+
+$`-8a = 5 \implies a = \dfrac{-5}{8} \quad (3)`$
+
+Sub $`3`$ into $`(2)`$:
+
+$`5 = \dfrac{-5}{8} + c \implies c = \dfrac{45}{8}`$
+
+$`\therefore`$ Our equation is $`y = \dfrac{-5}{8}(x-3)^2 + \dfrac{45}{8}`$