48t=16t^2+20

Solution for 48t=16t^2+20 equation:

48t=16t^2+20

We move all terms to the left:

48t-(16t^2+20)=0

We get rid of parentheses

-16t^2+48t-20=0

a = -16; b = 48; c = -20;
Δ = b2-4ac
Δ = 482-4·(-16)·(-20)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-32}{2*-16}=\frac{-80}{-32}
=2+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+32}{2*-16}=\frac{-16}{-32}
=1/2
$