User:Rosuav/Sandbox

Random sandbox location for user Rosuav. You don't want to bother with it.

Old MacDonald had a math, 2.718, $\sqrt{-1}$, 2.718, $\sqrt{-1}$, 0. (See? I told you you didn't want to bother with this page...)

$$7x+2 \overline{)28x^3-48x^2-30x-4}$$


 * $$4x^2\,$$

$$7x+2 \overline{)28x^3-48x^2-30x-4}$$
 * $$\,\,\,\,\dfrac{28x^3+8x^2}{\,\,\,\,\,-56x^2}$$

$$\overline{k} = \sum_{i=1}^n P(k_i)k_i$$

$$x^3 - 9x^{3/2} + 8 = 0$$

$$(x^{3/2})^2 - 9(x^{3/2}) + 8 = 0$$

$$y = x^{3/2}$$ $$(y)^2 - 9(y) + 8 = 0$$

$$y = 1: (1)^2 - 9(1) + 8 = 0$$ $$y = 8: (8)^2 - 9(8) + 8 = 0$$

$$x = y^{2/3}$$ $$x = 1^{2/3} = 1$$ $$x = 8^{2/3} = 4$$

$$\dfrac{x^3}{5} + 17 = 19$$

$$x^{3/5} + 17 = 19$$

$$\dfrac{\sqrt[3]{2m+3}}{-5} + 2 = 3$$

Simplifying $$\dfrac{2x}{x-3} + \dfrac{4}{x+2} = \dfrac{-2}{(x-3)*(x+2)}$$

$$\dfrac{(x+2)(2x)}{x-3} + \dfrac{(x+2)(4)}{x+2} = \dfrac{(x+2)(-2)}{(x-3)(x+2)}$$

$$\dfrac{(x+2)(2x)}{x-3} + 4 = \dfrac{(-2)}{(x-3)}$$

$$\dfrac{(x+2)(2x)(x-3)}{x-3} + 4(x-3) = \dfrac{(-2)(x-3)}{(x-3)}$$

$$(x+2)(2x) + 4(x-3) = -2$$

$$2x^2+4x + 4x-4(3) + 2 = 0$$

$$2x^2 + 8x - 10 = 0$$

$$x^2 + 4x - 5 = 0$$