12t^2+20t-48=0

Solution for 12t^2+20t-48=0 equation:

12t^2+20t-48=0

a = 12; b = 20; c = -48;
Δ = b2-4ac
Δ = 202-4·12·(-48)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-52}{2*12}=\frac{-72}{24}
=-3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+52}{2*12}=\frac{32}{24}
=1+1/3
$