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What is 3t^2-13t=10?

 

 

If I remember

3t^2-13t+10=0

(3t+5)(t-2)

3t+5=0  t-2=0   

    -5 -5 +2 +2

3t=-5     t=2

t=-5/3

 

My answer came out wrong, but I don't know what I did wrong.  Can someone refresh my memory? 

Guest Aug 30, 2017

Best Answer 

 #1
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The formula you are lookin for is:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 

In your case: \(3t^2 - 13t - 10 = 0\)  

so: \(a = 3, b = -13, c = -10\)

and you can just calculate x using this coefficients. 

Alternatively, you could separate equation using factors (as you tried to do, I think):
\(3t^2 - 13t - 10 = 0 \\ 3t^2 - 15t + 2t - 10 = 0\\ 3t(t-5) + 2(t-5) = 0\\ (t-5)(3t+2)=0\\\)

which yields two solutions:

\(t_1=5 , \\ 3t_2 = -2 \Rightarrow t_2 = -\frac{2}{5}\)

 

Cheers. 

Guest Aug 30, 2017
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1+0 Answers

 #1
avatar
0
Best Answer

 

The formula you are lookin for is:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 

In your case: \(3t^2 - 13t - 10 = 0\)  

so: \(a = 3, b = -13, c = -10\)

and you can just calculate x using this coefficients. 

Alternatively, you could separate equation using factors (as you tried to do, I think):
\(3t^2 - 13t - 10 = 0 \\ 3t^2 - 15t + 2t - 10 = 0\\ 3t(t-5) + 2(t-5) = 0\\ (t-5)(3t+2)=0\\\)

which yields two solutions:

\(t_1=5 , \\ 3t_2 = -2 \Rightarrow t_2 = -\frac{2}{5}\)

 

Cheers. 

Guest Aug 30, 2017

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