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

2 Answers
Jul 28, 2018

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

Jul 28, 2018

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