S(t)=-t^2+17t-42

Solution for S(t)=-t^2+17t-42 equation:

(S)=-S^2+17S-42

We move all terms to the left:

(S)-(-S^2+17S-42)=0

We get rid of parentheses

S^2-17S+S+42=0

We add all the numbers together, and all the variables

S^2-16S+42=0

a = 1; b = -16; c = +42;
Δ = b2-4ac
Δ = -162-4·1·42
Δ = 88
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{22}}{2*1}=\frac{16-2\sqrt{22}}{2}
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{22}}{2*1}=\frac{16+2\sqrt{22}}{2}
$