Problem of the Week

Updated at Jan 29, 2024 3:57 PM

How can we solve the equation t25−t+22=15\frac{{t}^{2}}{5}-\frac{t+2}{2}=\frac{1}{5}?

Below is the solution.



t25−t+22=15\frac{{t}^{2}}{5}-\frac{t+2}{2}=\frac{1}{5}

1
Multiply both sides by 1010 (the LCM of 5,25, 2).
2t2−5(t+2)=22{t}^{2}-5(t+2)=2

2
Expand.
2t2−5t−10=22{t}^{2}-5t-10=2

3
Move all terms to one side.
2t2−5t−10−2=02{t}^{2}-5t-10-2=0

4
Simplify  2t2−5t−10−22{t}^{2}-5t-10-2  to  2t2−5t−122{t}^{2}-5t-12.
2t2−5t−12=02{t}^{2}-5t-12=0

5
Split the second term in 2t2−5t−122{t}^{2}-5t-12 into two terms.
2t2+3t−8t−12=02{t}^{2}+3t-8t-12=0

6
Factor out common terms in the first two terms, then in the last two terms.
t(2t+3)−4(2t+3)=0t(2t+3)-4(2t+3)=0

7
Factor out the common term 2t+32t+3.
(2t+3)(t−4)=0(2t+3)(t-4)=0

8
Solve for tt.
t=−32,4t=-\frac{3}{2},4

Done

Decimal Form: -1.5, 4
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