-4=2t-8t^2

Solution for -4=2t-8t^2 equation:

-4=2t-8t^2

We move all terms to the left:

-4-(2t-8t^2)=0

We get rid of parentheses

8t^2-2t-4=0

a = 8; b = -2; c = -4;
Δ = b2-4ac
Δ = -22-4·8·(-4)
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{33}}{2*8}=\frac{2-2\sqrt{33}}{16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{33}}{2*8}=\frac{2+2\sqrt{33}}{16}
$