$$\frac{1}{2}h+\frac{1}{3}h(h-6) = \frac{5}{6}h+2$$
Answer
$$ \begin{matrix}h_1 = \dfrac{ 7 }{ 2 }-\dfrac{\sqrt{ 73 }}{ 2 } & h_2 = \dfrac{ 7 }{ 2 }+\dfrac{\sqrt{ 73 }}{ 2 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1}{2}h+\frac{1}{3}h(h-6) &= \frac{5}{6}h+2&& \text{multiply ALL terms by } \color{blue}{ 6 }. \\[1 em]6 \cdot \frac{1}{2}h+6\frac{1}{3}h(h-6) &= 6 \cdot \frac{5}{6}h+6\cdot2&& \text{cancel out the denominators} \\[1 em]3h+2h^2-12h &= 5h+12&& \text{simplify left side} \\[1 em]2h^2-9h &= 5h+12&& \text{move all terms to the left hand side } \\[1 em]2h^2-9h-5h-12 &= 0&& \text{simplify left side} \\[1 em]2h^2-14h-12 &= 0&& \\[1 em] \end{aligned} $$
$ 2x^{2}-14x-12 = 0 $ is a quadratic equation.
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