4(3t^2+2t-8)=0

Solution for 4(3t^2+2t-8)=0 equation:

4(3t^2+2t-8)=0

We multiply parentheses

12t^2+8t-32=0

a = 12; b = 8; c = -32;
Δ = b2-4ac
Δ = 82-4·12·(-32)
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*12}=\frac{-48}{24}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+40}{2*12}=\frac{32}{24}
=1+1/3
$