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

2 Answers
Shivang Madaan
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#

maganbhai P.
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#

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