Problem of the Week

Updated at Feb 26, 2024 9:23 AM

This week we have another equation problem:

How would you solve \(\frac{4(t-3)(2+t)}{3}=\frac{56}{3}\)?

Let's start!



\[\frac{4(t-3)(2+t)}{3}=\frac{56}{3}\]

1
Multiply both sides by \(3\).
\[4(t-3)(2+t)=56\]

2
Expand.
\[8t+4{t}^{2}-24-12t=56\]

3
Simplify  \(8t+4{t}^{2}-24-12t\)  to  \(-4t+4{t}^{2}-24\).
\[-4t+4{t}^{2}-24=56\]

4
Move all terms to one side.
\[4t-4{t}^{2}+24+56=0\]

5
Simplify  \(4t-4{t}^{2}+24+56\)  to  \(4t-4{t}^{2}+80\).
\[4t-4{t}^{2}+80=0\]

6
Factor out the common term \(4\).
\[4(t-{t}^{2}+20)=0\]

7
Factor out the negative sign.
\[4\times -({t}^{2}-t-20)=0\]

8
Divide both sides by \(4\).
\[-{t}^{2}+t+20=0\]

9
Multiply both sides by \(-1\).
\[{t}^{2}-t-20=0\]

10
Factor \({t}^{2}-t-20\).
\[(t-5)(t+4)=0\]

11
Solve for \(t\).
\[t=5,-4\]

Done
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