400=16t^2+3t

Solution for 400=16t^2+3t equation:

400=16t^2+3t

We move all terms to the left:

400-(16t^2+3t)=0

We get rid of parentheses

-16t^2-3t+400=0

a = -16; b = -3; c = +400;
Δ = b2-4ac
Δ = -32-4·(-16)·400
Δ = 25609
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{-(-3)-\sqrt{25609}}{2*-16}=\frac{3-\sqrt{25609}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{25609}}{2*-16}=\frac{3+\sqrt{25609}}{-32}
$