H(t)=-5t^2+14t+2

Solution for H(t)=-5t^2+14t+2 equation:

(H)=-5H^2+14H+2

We move all terms to the left:

(H)-(-5H^2+14H+2)=0

We get rid of parentheses

5H^2-14H+H-2=0

We add all the numbers together, and all the variables

5H^2-13H-2=0

a = 5; b = -13; c = -2;
Δ = b2-4ac
Δ = -132-4·5·(-2)
Δ = 209
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{-(-13)-\sqrt{209}}{2*5}=\frac{13-\sqrt{209}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{209}}{2*5}=\frac{13+\sqrt{209}}{10}
$