0(t)=-16t^2+112t+25

Solution for 0(t)=-16t^2+112t+25 equation:

0(t)=-16t^2+112t+25

We move all terms to the left:

0(t)-(-16t^2+112t+25)=0

We add all the numbers together, and all the variables

-(-16t^2+112t+25)+t=0

We get rid of parentheses

16t^2-112t+t-25=0

We add all the numbers together, and all the variables

16t^2-111t-25=0

a = 16; b = -111; c = -25;
Δ = b2-4ac
Δ = -1112-4·16·(-25)
Δ = 13921
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-111)-\sqrt{13921}}{2*16}=\frac{111-\sqrt{13921}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-111)+\sqrt{13921}}{2*16}=\frac{111+\sqrt{13921}}{32}
$