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