# Problem of the Week

## Updated at Dec 21, 2020 2:08 PM

This week we have another equation problem:

How would you solve $$t(4t-3)=27$$?

Let's start!

$t(4t-3)=27$

 1 Expand.$4{t}^{2}-3t=27$2 Move all terms to one side.$4{t}^{2}-3t-27=0$3 Split the second term in $$4{t}^{2}-3t-27$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$4\times -27=-108$2 Ask: Which two numbers add up to $$-3$$ and multiply to $$-108$$?$$9$$ and $$-12$$3 Split $$-3t$$ as the sum of $$9t$$ and $$-12t$$.$4{t}^{2}+9t-12t-27$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$4{t}^{2}+9t-12t-27=0$4 Factor out common terms in the first two terms, then in the last two terms.$t(4t+9)-3(4t+9)=0$5 Factor out the common term $$4t+9$$.$(4t+9)(t-3)=0$6 Solve for $$t$$.1 Ask: When will $$(4t+9)(t-3)$$ equal zero?When $$4t+9=0$$ or $$t-3=0$$2 Solve each of the 2 equations above.$t=-\frac{9}{4},3$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$t=-\frac{9}{4},3$DoneDecimal Form: -2.25, 3t=-9/4,3