First, factor a #t# from each term on the left giving:
#t(10 - 2t) = 0#
Now, solve each term on the left for #0# to find the values of #t#:
Solution 1:
#t = 0#
Solution 2:
#10 - 2t = 0#
#-color(red)(10) + 10 - 2t = -color(red)(10) + 0#
#0 - 2t = -10#
#-2t = -10#
#(-2t)/color(red)(-2) = -10/color(red)(-2)#
#(color(red)(cancel(color(black)(-2)))t)/cancel(color(red)(-2)) = 5#
#t = 5#
**The Solutions Are: #t = 0# and #t = 5#
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First, rewrite this equation in standard form:
#-2t^2 + 10t + 0 = 0#
We can now use the quadratic equation to solve this problem:
The quadratic formula states:
For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:
#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))#
Substituting:
#color(red)(-2)# for #color(red)(a)#
#color(blue)(10)# for #color(blue)(b)#
#color(green)(0)# for #color(green)(c)# gives:
#t = (-color(blue)(10) +- sqrt(color(blue)(10)^2 - (4 * color(red)(-2) * color(green)(0))))/(2 * color(red)(-2))#
#t = (-color(blue)(10) +- sqrt(100 - 0))/(-4)#
#t = (-color(blue)(10) +- sqrt(100))/(-4)#
#t = (-color(blue)(10) - sqrt(100))/(-4)# and #t = (-color(blue)(10) + sqrt(100))/(-4)#
#t = (-color(blue)(10) - 10)/(-4)# and #t = (-color(blue)(10) + 10)/(-4)#
#t = (-20)/(-4)# and #t = 0/(-4)#
#t = 5# and #t = 0#
