Question

How do you solve `10t^2 - 29t = -10`?

Answer 1

`t=2/5 or 5/2`

Explanation:

`10t^2-29t+10=0`
you can solve it by 2 methods, the quadratic formula or splitting the middle term.
I am doing it by the latter one.

the product of roots is 100 and the sum is -29
the suitable pair will be -25 and -4
therefore,
`10t^2-25t-4t+10=0`
`5t(2t-5)-2(2t-5)=0`
`(5t-2)(2t-5)=0`
which implies that,
`t=2/5 or 5/2`

Answer 2

`t=5/2 or t=2/5`

Explanation:

Here,

`10t^2-29t=-10`

`10t^2-29t+10=0`

`10t^2-25t-4t+10=0`

`5t(2t-5)-2(2t-5)=0`

`(2t-5)(5t-2)=0`

`2t-5=0or5t-2=0`

`2t=5 or5t=2`

`t=5/2 or t=2/5`

...........................................OR......................................

`10t^2-29t+10=0`

Comparing with `at^2+bt+c=0`

`a=10, b=-29 and c=10`

So,

`t=(-b+-sqrt(b^2-4ac))/(2a)`

`:.t=(29+-sqrt(29^2-4(10)(10)))/(2xx10)`

`:.t=(29+-sqrt(841-400))/20`

`:.t=(29+-21)/20`

`:.t=(29+21)/20 or t=(29-21)/20`

`:.t=50/20 or t=8/20`

`:.t=5/2 or t=2/5`