How do you solve using the quadratic formula for #h(t) = -0.5t^2 + 10t + 22#?

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Answer 1 · 2018-03-14T02:49:50

Using the quadratic formula, the solutions for #ax^2+bx+c=0# are

#x=(-b+-sqrt(b^2-4ac))/(2a)#

So, we got here:

#a=-0.5,b=10,c=22#

Therefore,

#t=(-10+-sqrt(100-4*-0.5*22))/-1#

#=(-10+-sqrt(100+44))/-1#

#=(-10+-sqrt(144))/-1#

#=(-10+-12)/-1#

#=-(-10+-12)#

#=-(2,-22)#

#=-2,22#

So, the solutions for #t# will be #tin{-2,22}#.