(R+ωL)(R-ωL)
= R^2-ωL

Z^2-R^2
= (Z+R)(Z-R)


h= -4.9t^2+55t+12
make ‘t’ subject

h= -4.9t^2+55t+12
h-12=-4.9t^2+55t+12
-12h=50.1t

-12h/50.1=t


Thank you

For the last one, you can't subtract a t term from a t^2 term:confused: .

You just broke the laws of algebra. :) :rolleyes: :p

You can easily check your answer by subbing it back in to the equation. If it works, you should get h. But you won't.:eek:

-\frac{49}{10}t^{2}+55t+12=h

Try completing the square. Can you do that?

Upon doing that you get \frac{-(49t-275)^{2}}{490}+\frac{16301}{98}

Multiply the right side by 5 to get the denominators the same, then we get:

\frac{-(49t-275)^{2}+81505}{490}=h

(49t-275)^{2}-81505=-490h

(49t-275)^{2}=-490h+81505

49t-275=\sqrt{-490h+81505}

t=\frac{\sqrt{-490h+81505}+275}{49}, \;\ \frac{-\sqrt{-490h+81505}-275}{49}

There, see what happened?