H(t)=-5t^2+10t+6

Solution for H(t)=-5t^2+10t+6 equation:

(H)=-5H^2+10H+6

We move all terms to the left:

(H)-(-5H^2+10H+6)=0

We get rid of parentheses

5H^2-10H+H-6=0

We add all the numbers together, and all the variables

5H^2-9H-6=0

a = 5; b = -9; c = -6;
Δ = b2-4ac
Δ = -92-4·5·(-6)
Δ = 201
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{201}}{2*5}=\frac{9-\sqrt{201}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{201}}{2*5}=\frac{9+\sqrt{201}}{10}
$