0=-16t^2-72t+280

Solution for 0=-16t^2-72t+280 equation:

0=-16t^2-72t+280

We move all terms to the left:

0-(-16t^2-72t+280)=0

We add all the numbers together, and all the variables

-(-16t^2-72t+280)=0

We get rid of parentheses

16t^2+72t-280=0

a = 16; b = 72; c = -280;
Δ = b2-4ac
Δ = 722-4·16·(-280)
Δ = 23104
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{23104}=152$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-152}{2*16}=\frac{-224}{32}
=-7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+152}{2*16}=\frac{80}{32}
=2+1/2
$