0(t)=-16t^2+56t

Solution for 0(t)=-16t^2+56t equation:

0(t)=-16t^2+56t

We move all terms to the left:

0(t)-(-16t^2+56t)=0

We add all the numbers together, and all the variables

-(-16t^2+56t)+t=0

We get rid of parentheses

16t^2-56t+t=0

We add all the numbers together, and all the variables

16t^2-55t=0

a = 16; b = -55; c = 0;
Δ = b2-4ac
Δ = -552-4·16·0
Δ = 3025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3025}=55$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-55)-55}{2*16}=\frac{0}{32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-55)+55}{2*16}=\frac{110}{32}
=3+7/16
$