-8(2t^2-t-3)=0

Solution for -8(2t^2-t-3)=0 equation:

-8(2t^2-t-3)=0

We multiply parentheses

-16t^2+8t+24=0

a = -16; b = 8; c = +24;
Δ = b2-4ac
Δ = 82-4·(-16)·24
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-40}{2*-16}=\frac{-48}{-32}
=1+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+40}{2*-16}=\frac{32}{-32}
=-1
$