0=-(16t^2-48t-8)

Solution for 0=-(16t^2-48t-8) equation:

0=-(16t^2-48t-8)

We move all terms to the left:

0-(-(16t^2-48t-8))=0

We add all the numbers together, and all the variables

-(-(16t^2-48t-8))=0


We calculate terms in parentheses: -(-(16t^2-48t-8)), so:

-(16t^2-48t-8)

We get rid of parentheses

-16t^2+48t+8

Back to the equation:

-(-16t^2+48t+8)


We get rid of parentheses

16t^2-48t-8=0

a = 16; b = -48; c = -8;
Δ = b2-4ac
Δ = -482-4·16·(-8)
Δ = 2816
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:

$\sqrt{\Delta}=\sqrt{2816}=\sqrt{256*11}=\sqrt{256}*\sqrt{11}=16\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16\sqrt{11}}{2*16}=\frac{48-16\sqrt{11}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16\sqrt{11}}{2*16}=\frac{48+16\sqrt{11}}{32}
$