$$\((sqrt{3x}-2)^2)<5\)$$
\(ax^2 + bx + c = 0\)
$$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$
$$\sqrt{\left(3x-2\right)^2}\lt 5$$
$$\left(\frac{\sqrt x}{y^3}\right)$$
$$\left\right$$
$$\sqrt 2x + 5 = 3$$
$$x + 2 – \frac {15} {x} = 0$$
$$16 = \frac {k} {11}$$
$$2y = x + 7$$
$$\int \left(x\right) = x^2 + 4x + 5$$
$$x = -2$$, or $$x=5$$
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