0=16t^2+21t-6

Solution for 0=16t^2+21t-6 equation:

0=16t^2+21t-6

We move all terms to the left:

0-(16t^2+21t-6)=0

We add all the numbers together, and all the variables

-(16t^2+21t-6)=0

We get rid of parentheses

-16t^2-21t+6=0

a = -16; b = -21; c = +6;
Δ = b2-4ac
Δ = -212-4·(-16)·6
Δ = 825
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:

$\sqrt{\Delta}=\sqrt{825}=\sqrt{25*33}=\sqrt{25}*\sqrt{33}=5\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-5\sqrt{33}}{2*-16}=\frac{21-5\sqrt{33}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+5\sqrt{33}}{2*-16}=\frac{21+5\sqrt{33}}{-32}
$