2(2t-1)(t+8)=0

Solution for 2(2t-1)(t+8)=0 equation:

2(2t-1)(t+8)=0

We multiply parentheses ..

2(+2t^2+16t-1t-8)=0

We multiply parentheses

4t^2+32t-2t-16=0

We add all the numbers together, and all the variables

4t^2+30t-16=0

a = 4; b = 30; c = -16;
Δ = b2-4ac
Δ = 302-4·4·(-16)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-34}{2*4}=\frac{-64}{8}
=-8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+34}{2*4}=\frac{4}{8}
=1/2
$