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)+4t2=133(t-3)+4{t}^{2}=13?

Here are the steps:



3(t−3)+4t2=133(t-3)+4{t}^{2}=13

1
Expand.
3t−9+4t2=133t-9+4{t}^{2}=13

2
Move all terms to one side.
3t−9+4t2−13=03t-9+4{t}^{2}-13=0

3
Simplify  3t−9+4t2−133t-9+4{t}^{2}-13  to  3t−22+4t23t-22+4{t}^{2}.
3t−22+4t2=03t-22+4{t}^{2}=0

4
Split the second term in 3t−22+4t23t-22+4{t}^{2} into two terms.
4t2+11t−8t−22=04{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)=0t(4t+11)-2(4t+11)=0

6
Factor out the common term 4t+114t+11.
(4t+11)(t−2)=0(4t+11)(t-2)=0

7
Solve for tt.
t=−114,2t=-\frac{11}{4},2

Done

Decimal Form: -2.75, 2
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