$x^2\:+7\:x\:+12\:=\:0$
$4\left(x+1\right)^{\frac{1}{2}}-5\left(x+1\right)^{\frac{3}{2}}$
$\frac{x^3-x+1}{x+1}$
$-x^2+3x-\frac{3}{2}^2$
$40+40+40+45$
$x^2=1x^2$
$3x+12$
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