Problem of the Week

Updated at Apr 29, 2024 12:01 PM

For this week we've brought you this equation problem.

How would you solve the equation \(3(t-3)+4{t}^{2}=13\)?

Here are the steps:



\[3(t-3)+4{t}^{2}=13\]

1
Expand.
\[3t-9+4{t}^{2}=13\]

2
Move all terms to one side.
\[3t-9+4{t}^{2}-13=0\]

3
Simplify  \(3t-9+4{t}^{2}-13\)  to  \(3t-22+4{t}^{2}\).
\[3t-22+4{t}^{2}=0\]

4
Split the second term in \(3t-22+4{t}^{2}\) into two terms.
\[4{t}^{2}+11t-8t-22=0\]

5
Factor out common terms in the first two terms, then in the last two terms.
\[t(4t+11)-2(4t+11)=0\]

6
Factor out the common term \(4t+11\).
\[(4t+11)(t-2)=0\]

7
Solve for \(t\).
\[t=-\frac{11}{4},2\]

Done

Decimal Form: -2.75, 2
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