0=16t^2+116t-28

Solution for 0=16t^2+116t-28 equation:

0=16t^2+116t-28

We move all terms to the left:

0-(16t^2+116t-28)=0

We add all the numbers together, and all the variables

-(16t^2+116t-28)=0

We get rid of parentheses

-16t^2-116t+28=0

a = -16; b = -116; c = +28;
Δ = b2-4ac
Δ = -1162-4·(-16)·28
Δ = 15248
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:

$\sqrt{\Delta}=\sqrt{15248}=\sqrt{16*953}=\sqrt{16}*\sqrt{953}=4\sqrt{953}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-116)-4\sqrt{953}}{2*-16}=\frac{116-4\sqrt{953}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-116)+4\sqrt{953}}{2*-16}=\frac{116+4\sqrt{953}}{-32}
$