-2t^2+20t+11=0

Solution for -2t^2+20t+11=0 equation:

-2t^2+20t+11=0

a = -2; b = 20; c = +11;
Δ = b2-4ac
Δ = 202-4·(-2)·11
Δ = 488
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{488}=\sqrt{4*122}=\sqrt{4}*\sqrt{122}=2\sqrt{122}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{122}}{2*-2}=\frac{-20-2\sqrt{122}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{122}}{2*-2}=\frac{-20+2\sqrt{122}}{-4}
$