0=-16t^2+228t+21

Solution for 0=-16t^2+228t+21 equation:

0=-16t^2+228t+21

We move all terms to the left:

0-(-16t^2+228t+21)=0

We add all the numbers together, and all the variables

-(-16t^2+228t+21)=0

We get rid of parentheses

16t^2-228t-21=0

a = 16; b = -228; c = -21;
Δ = b2-4ac
Δ = -2282-4·16·(-21)
Δ = 53328
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{53328}=\sqrt{16*3333}=\sqrt{16}*\sqrt{3333}=4\sqrt{3333}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-228)-4\sqrt{3333}}{2*16}=\frac{228-4\sqrt{3333}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-228)+4\sqrt{3333}}{2*16}=\frac{228+4\sqrt{3333}}{32}
$