S(t)=-16t^2+48t+42

Solution for S(t)=-16t^2+48t+42 equation:

(S)=-16S^2+48S+42

We move all terms to the left:

(S)-(-16S^2+48S+42)=0

We get rid of parentheses

16S^2-48S+S-42=0

We add all the numbers together, and all the variables

16S^2-47S-42=0

a = 16; b = -47; c = -42;
Δ = b2-4ac
Δ = -472-4·16·(-42)
Δ = 4897
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-47)-\sqrt{4897}}{2*16}=\frac{47-\sqrt{4897}}{32}
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+\sqrt{4897}}{2*16}=\frac{47+\sqrt{4897}}{32}
$