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

Solution for 0=-4(4t^2-8t-2) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -(-4(4t^2-8t-2)), so:

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

We multiply parentheses

-16t^2+32t+8

Back to the equation:

-(-16t^2+32t+8)


We get rid of parentheses

16t^2-32t-8=0

a = 16; b = -32; c = -8;
Δ = b2-4ac
Δ = -322-4·16·(-8)
Δ = 1536
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:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16\sqrt{6}}{2*16}=\frac{32-16\sqrt{6}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16\sqrt{6}}{2*16}=\frac{32+16\sqrt{6}}{32}
$