21=5t+16t^2

Solution for 21=5t+16t^2 equation:

21=5t+16t^2

We move all terms to the left:

21-(5t+16t^2)=0

We get rid of parentheses

-16t^2-5t+21=0

a = -16; b = -5; c = +21;
Δ = b2-4ac
Δ = -52-4·(-16)·21
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-37}{2*-16}=\frac{-32}{-32}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+37}{2*-16}=\frac{42}{-32}
=-1+5/16
$