48=-4t^2+24t+28

Solution for 48=-4t^2+24t+28 equation:

48=-4t^2+24t+28

We move all terms to the left:

48-(-4t^2+24t+28)=0

We get rid of parentheses

4t^2-24t-28+48=0

We add all the numbers together, and all the variables

4t^2-24t+20=0

a = 4; b = -24; c = +20;
Δ = b2-4ac
Δ = -242-4·4·20
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-16}{2*4}=\frac{8}{8}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+16}{2*4}=\frac{40}{8}
=5
$