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

Question

3150 views around the world

You can reuse this answer
Creative Commons License

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 $tin{-2,22}$ .

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