H=11+10t-2t^2

Solution for H=11+10t-2t^2 equation:

=11+10H-2H^2

We move all terms to the left:

-(11+10H-2H^2)=0

We get rid of parentheses

2H^2-10H-11=0

a = 2; b = -10; c = -11;
Δ = b2-4ac
Δ = -102-4·2·(-11)
Δ = 188
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{188}=\sqrt{4*47}=\sqrt{4}*\sqrt{47}=2\sqrt{47}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{47}}{2*2}=\frac{10-2\sqrt{47}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{47}}{2*2}=\frac{10+2\sqrt{47}}{4}
$