H(t)=-4t^2+10

Solution for H(t)=-4t^2+10 equation:

(H)=-4H^2+10

We move all terms to the left:

(H)-(-4H^2+10)=0

We get rid of parentheses

4H^2+H-10=0

a = 4; b = 1; c = -10;
Δ = b2-4ac
Δ = 12-4·4·(-10)
Δ = 161
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{161}}{2*4}=\frac{-1-\sqrt{161}}{8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{161}}{2*4}=\frac{-1+\sqrt{161}}{8}
$