$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$

$(a+b)^2$

$$\eqalign{ (a+b)^2 &= (a+b)(a+b) \\ &= a^2 + ab + ba + b^2 \\ &= a^2 + 2ab + b^2 }$$

$(a-b)^2$

$$\eqalign{ (a-b)^2 &= (a-b)(a-b) \\ &= a^2 - ab - ba + b^2 \\ &= a^2 - 2ab + b^2 }$$

$(a-b)(a+b)$

$$\eqalign{ (a+b)(a-b) &= a^2 - ab + ba - b^2 \\ &= a^2 - b^2 }$$

$$ f(x) = {\sqrt{5x^2+2x-1}+(x-2)^2 } $$

$$ g(x) = {\frac{a}{1-a^2} }$$

$$ f(x) = {(x + 2) \over (2x + 1)} $$

$$ f(x) = { \sqrt[3]{x^2} }$$

$$ \sqrt[5]{34}$$

$$ cos^2 \theta + sin^2 \theta = 1 $$

$$ tan 2 \theta = {2tan \theta \over 1 - tan^2 \theta} $$

$$\eqalign{ cos 2 \theta = cos^2 \theta - sin^2 \theta \\ &= 2 cos^2 \theta -1 \\ &= 1 - 2sin^2 \theta }$$

Prove $ \sqrt{ 1 - cos^2 \theta \over 1- sin^2 \theta} = tan \theta $

$$ \sqrt{ 1 - cos^2 \theta \over 1- sin^2 \theta} = \sqrt{ sin^2 \theta \over cos^2 \theta} = {sin \theta \over cos \theta} = Tan \theta $$

$$\eqalign{ f(x) = {3x^4} \implies {dy \over dx} = 12x^3 }$$

$$\eqalign{ f(x) = {2x^{-3/2}} \implies {dy \over dx} = -3x^{-5/2} &= - {3 \over \sqrt{x^5}} }$$

If $x = 2t + 1$ and $ y = t^2$ find ${dy \over dx}$ ?

$$\eqalign{ x = 2t + 1 \implies {dx \over dt} = 2 \\ y = t^2 \implies {dy \over dt} = 2t \\ {dy \over dx} = {dy \over dt} \div {dx \over dt} \\ \implies 2t \div 2 = t }$$

Evaluate $\int_1^2 (x + 4)^2 dx $

$$\eqalign{ \int_1^2 (x + 4)^2 dx = \int_1^2 (x^2 + 8x + 16) dx \\ &= \left\lbrack {x^3 \over 3} + {8x^2 \over 2} + 16x \right\rbrack_1^2 \\ &= \left\lbrack {8 \over 3} + {8 * 4 \over 2} + 16 * 2 \right\rbrack - \left\lbrack {1 \over 3} + {8 \over 2} + 16 \right\rbrack }$$

$$ {\left\lbrack \matrix{2 & 3 \cr 4 & 5} \right\rbrack} * \left\lbrack \matrix{1 & 0 \cr 0 & 1} \right\rbrack = \left\lbrack \matrix{2 & 3 \cr 4 & 5} \right\rbrack $$

$$\sum_{n=1}^n n = {n \over 2} (n + 1) $$

$$\sum_{n=1}^n n^2 = {n \over 6} (n + 1)(2n + 1) $$

https://github.blog/changelog/2022-05-19-render-mathematical-expressions-in-markdown/

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