H(t)=-3t^2+14t+5

Solution for H(t)=-3t^2+14t+5 equation:

(H)=-3H^2+14H+5

We move all terms to the left:

(H)-(-3H^2+14H+5)=0

We get rid of parentheses

3H^2-14H+H-5=0

We add all the numbers together, and all the variables

3H^2-13H-5=0

a = 3; b = -13; c = -5;
Δ = b2-4ac
Δ = -132-4·3·(-5)
Δ = 229
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{229}}{2*3}=\frac{13-\sqrt{229}}{6}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{229}}{2*3}=\frac{13+\sqrt{229}}{6}
$