0=-4t^2+16t+48

Solution for 0=-4t^2+16t+48 equation:

0=-4t^2+16t+48

We move all terms to the left:

0-(-4t^2+16t+48)=0

We add all the numbers together, and all the variables

-(-4t^2+16t+48)=0

We get rid of parentheses

4t^2-16t-48=0

a = 4; b = -16; c = -48;
Δ = b2-4ac
Δ = -162-4·4·(-48)
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{-(-16)-32}{2*4}=\frac{-16}{8}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+32}{2*4}=\frac{48}{8}
=6
$