48-20t=12t^2

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

48-20t=12t^2

We move all terms to the left:

48-20t-(12t^2)=0

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