10=-16(t)^2+(30)(t)+9

Solution for 10=-16(t)^2+(30)(t)+9 equation:

10=-16(t)^2+(30)(t)+9

We move all terms to the left:

10-(-16(t)^2+(30)(t)+9)=0

We get rid of parentheses

16t^2-30t-9+10=0

We add all the numbers together, and all the variables

16t^2-30t+1=0

a = 16; b = -30; c = +1;
Δ = b2-4ac
Δ = -302-4·16·1
Δ = 836
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:

$\sqrt{\Delta}=\sqrt{836}=\sqrt{4*209}=\sqrt{4}*\sqrt{209}=2\sqrt{209}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-2\sqrt{209}}{2*16}=\frac{30-2\sqrt{209}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{209}}{2*16}=\frac{30+2\sqrt{209}}{32}
$