H(t)=-3t2+1

Solution for H(t)=-3t2+1 equation:

(H)=-3H^2+1

We move all terms to the left:

(H)-(-3H^2+1)=0

We get rid of parentheses

3H^2+H-1=0

a = 3; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·3·(-1)
Δ = 13
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{-(1)-\sqrt{13}}{2*3}=\frac{-1-\sqrt{13}}{6}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{13}}{2*3}=\frac{-1+\sqrt{13}}{6}
$