0=-16t2+-18t+210

Solution for 0=-16t2+-18t+210 equation:

0=-16t^2+-18t+210

We move all terms to the left:

0-(-16t^2+-18t+210)=0

We add all the numbers together, and all the variables

-(-16t^2+-18t+210)=0

We use the square of the difference formula

-(-16t^2-18t+210)=0

We get rid of parentheses

16t^2+18t-210=0

a = 16; b = 18; c = -210;
Δ = b2-4ac
Δ = 182-4·16·(-210)
Δ = 13764
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:

$\sqrt{\Delta}=\sqrt{13764}=\sqrt{4*3441}=\sqrt{4}*\sqrt{3441}=2\sqrt{3441}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{3441}}{2*16}=\frac{-18-2\sqrt{3441}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{3441}}{2*16}=\frac{-18+2\sqrt{3441}}{32}
$