User:Rosuav/Sandbox

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

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

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

$4x^{2}\,$

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

$\,\,\,\,{\dfrac {28x^{3}+8x^{2}}{\,\,\,\,\,-56x^{2}}}$

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${\overline {k}}=\sum _{i=1}^{n}P(k_{i})k_{i}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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