-8t2+8t+12=8

Solution for -8t2+8t+12=8 equation:

-8t^2+8t+12=8

We move all terms to the left:

-8t^2+8t+12-(8)=0

We add all the numbers together, and all the variables

-8t^2+8t+4=0

a = -8; b = 8; c = +4;
Δ = b2-4ac
Δ = 82-4·(-8)·4
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{3}}{2*-8}=\frac{-8-8\sqrt{3}}{-16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{3}}{2*-8}=\frac{-8+8\sqrt{3}}{-16}
$