S=18-11t+t2

Solution for S=18-11t+t2 equation:

=18-11S+S2

We move all terms to the left:

-(18-11S+S2)=0

We add all the numbers together, and all the variables

-(-11S+S^2+18)=0

We get rid of parentheses

-S^2+11S-18=0

We add all the numbers together, and all the variables

-1S^2+11S-18=0

a = -1; b = 11; c = -18;
Δ = b2-4ac
Δ = 112-4·(-1)·(-18)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-7}{2*-1}=\frac{-18}{-2}
=+9
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+7}{2*-1}=\frac{-4}{-2}
=+2
$