0=-16t^2+88t+12

Solution for 0=-16t^2+88t+12 equation:

0=-16t^2+88t+12

We move all terms to the left:

0-(-16t^2+88t+12)=0

We add all the numbers together, and all the variables

-(-16t^2+88t+12)=0

We get rid of parentheses

16t^2-88t-12=0

a = 16; b = -88; c = -12;
Δ = b2-4ac
Δ = -882-4·16·(-12)
Δ = 8512
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:

$\sqrt{\Delta}=\sqrt{8512}=\sqrt{64*133}=\sqrt{64}*\sqrt{133}=8\sqrt{133}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-88)-8\sqrt{133}}{2*16}=\frac{88-8\sqrt{133}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-88)+8\sqrt{133}}{2*16}=\frac{88+8\sqrt{133}}{32}
$