Solution 18.7a
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(New page: <math>\begin{align} & 3{{t}^{2}}+20t=500 \\ & 3{{t}^{2}}+20t-500=0 \\ & (3t+50)(t-10)=0 \\ & t=-\frac{50}{3}\text{ or }t=10 \\ & \text{It is at a height of 500 m when }t=10 \text { s...)
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\displaystyle \begin{align} & 3{{t}^{2}}+20t=500 \\ & 3{{t}^{2}}+20t-500=0 \\ & (3t+50)(t-10)=0 \\ & t=-\frac{50}{3}\text{ or }t=10 \\ & \text{It is at a height of 500 m when }t=10 \text { s}, \\ & \text{(the negative solution has no physical meaning). }\\ \end{align}
