S=-2t^2+14t+11

Solution for S=-2t^2+14t+11 equation:

=-2S^2+14S+11

We move all terms to the left:

-(-2S^2+14S+11)=0

We get rid of parentheses

2S^2-14S-11=0

a = 2; b = -14; c = -11;
Δ = b2-4ac
Δ = -142-4·2·(-11)
Δ = 284
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:

$\sqrt{\Delta}=\sqrt{284}=\sqrt{4*71}=\sqrt{4}*\sqrt{71}=2\sqrt{71}$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{71}}{2*2}=\frac{14-2\sqrt{71}}{4}
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{71}}{2*2}=\frac{14+2\sqrt{71}}{4}
$