16t^2+64t+21=0

Solution for 16t^2+64t+21=0 equation:

16t^2+64t+21=0

a = 16; b = 64; c = +21;
Δ = b2-4ac
Δ = 642-4·16·21
Δ = 2752
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:

$\sqrt{\Delta}=\sqrt{2752}=\sqrt{64*43}=\sqrt{64}*\sqrt{43}=8\sqrt{43}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-8\sqrt{43}}{2*16}=\frac{-64-8\sqrt{43}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+8\sqrt{43}}{2*16}=\frac{-64+8\sqrt{43}}{32}
$