0=-16t^2+t+225

Solution for 0=-16t^2+t+225 equation:

0=-16t^2+t+225

We move all terms to the left:

0-(-16t^2+t+225)=0

We add all the numbers together, and all the variables

-(-16t^2+t+225)=0

We get rid of parentheses

16t^2-t-225=0

We add all the numbers together, and all the variables

16t^2-1t-225=0

a = 16; b = -1; c = -225;
Δ = b2-4ac
Δ = -12-4·16·(-225)
Δ = 14401
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{-(-1)-\sqrt{14401}}{2*16}=\frac{1-\sqrt{14401}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{14401}}{2*16}=\frac{1+\sqrt{14401}}{32}
$