1+20t-5t^2=11

Solution for 1+20t-5t^2=11 equation:

1+20t-5t^2=11

We move all terms to the left:

1+20t-5t^2-(11)=0

We add all the numbers together, and all the variables

-5t^2+20t-10=0

a = -5; b = 20; c = -10;
Δ = b2-4ac
Δ = 202-4·(-5)·(-10)
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{2}}{2*-5}=\frac{-20-10\sqrt{2}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{2}}{2*-5}=\frac{-20+10\sqrt{2}}{-10}
$