MaxWong

Good answer Guest #1.

Example: If you have x^2 = 25......

$x^2=25\\ x=\pm\sqrt{25}\\ \;\;=\pm 5\\ \therefore\text{ x = 5 or x = -5}$

Don't be lazy guys, it's an easy one!! :D

$\dfrac{4}{7}+\dfrac{3}{14}\\ = \dfrac{8}{14}+\dfrac{3}{14}\\ =\dfrac{11}{14}$

$\frac{7}{3}=\frac{4}{x}\\ 7x=12\\ x=\frac{12}{7}$

$24-32\div 4 + 6 = 22$

Guest #1 don't be so rude.

$\dfrac{4|y+3|+12}{4}\\ =|y+3|+3$

x - 12/6 = 9

x - 2 = 9

x = 11

Or you mean (x-12)/6=9?

(x-12)/6 = 9

x-12 = 54

x = 66

$(6^{-4})^{-6}\\ = 6^{-4\times -6}\\ =6^{24}$

$\dfrac{1+3i}{2-5i}\\ =\dfrac{(1+3i)(2+5i)}{(2-5i)(2+5i)}\\ =\dfrac{-13+11i}{29}\\ =-\dfrac{13}{29}+\dfrac{11i}{29}$

Given numbers? Equation? Which numbers and equation= ?

For me, I usually use factorization or quadratic formula. 

Factorization example:

$x(x-5)=-4\\ x^2-5x+4=0\\ \color{red}(x-1)\color{aqua}(x-4)\color{black}=0\\ \text{Set each factor to 0.}\\ \color{red}x-1=0\\ \mathbf{\color{red}x=1}\\ \color{aqua}x-4=0\\ \mathbf{\color{aqua}x=4}\\ \therefore \color{red}x=1 \color{black} \text{ and } \color{aqua}x=4$

Quadratic formula example: 

$x^2-x-1=0\\ x=\dfrac{-(-1)\pm\sqrt{(-1)^2-(4)(1)(-1)}}{2(1)}\\ \;\;=\dfrac{1\pm\sqrt5}{2}\\ \therefore x=\dfrac{1+\sqrt5}{2}\text{ and }x=\dfrac{1-\sqrt5}{2}$

$\text{Just for you to know: The solution of }ax^2+bx+c=0\\\text{ is }x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

Complementation of the square example: (I usually call it completing the square)

$x^2+8x+15=0\\ x^2+8x+16-1=0\\ (x+4)^2=1\\ x+4=\pm \sqrt1\\ \;\;\;\;\;\;\;\;\;\!=\pm 1\\ x= -3\text{ and }x=-5$