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10t^2+5t=5t^2+11t+8

 Sep 3, 2015
 #1
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10t^2+5t=5t^2+11t+8

 

\(10t^2+5t=5t^2+11t+8\\ 10t^2-5t^2+5t-11t-8=0\\ 5t^2-6t-8=0\\ \)

Now you could use the quadratic formula but I am going to factorise it.

I need 2 numbers that multiply to give 5*-8=-40   and add to give -6

-10 and +4 work

change -6t into  -10t+4t

 

\(5t^2-10t+4t-8=0\\ 5t(t-2)+4(t-2)=0\\ (5t+4)(t-2)=0\\ 5t+4=0 \qquad or \qquad t-2=0\\ 5t=-4 \qquad or \qquad t=2\\ t=-\frac{4}{5} \qquad or \qquad t=2\\ \)

 Sep 3, 2015

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