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

Question

Hazel Anglin · Accepted Answer

#t=2/5 or 5/2# #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#

Savannah Anderson · Answer

#t=5/2 or t=2/5#

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#

Ethan Adkins · Answer

undefined To solve the equation $10t^2 - 29t = -10$, you can rearrange it into the standard quadratic form $10t^2 - 29t + 10 = 0$, and then apply the quadratic formula: $t = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$, where $a = 10$, $b = -29$, and $c = 10$. Calculate the discriminant $b^2 - 4ac$, and then substitute the values into the quadratic formula to find the solutions for $t$.