21=7+38t-16t^2=0

Solution for 21=7+38t-16t^2=0 equation:

21=7+38t-16t^2=0

We move all terms to the left:

21-(7+38t-16t^2)=0

We get rid of parentheses

16t^2-38t-7+21=0

We add all the numbers together, and all the variables

16t^2-38t+14=0

a = 16; b = -38; c = +14;
Δ = b2-4ac
Δ = -382-4·16·14
Δ = 548
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:

$\sqrt{\Delta}=\sqrt{548}=\sqrt{4*137}=\sqrt{4}*\sqrt{137}=2\sqrt{137}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{137}}{2*16}=\frac{38-2\sqrt{137}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{137}}{2*16}=\frac{38+2\sqrt{137}}{32}
$