H(t)=-2t^2+10t+3

Solution for H(t)=-2t^2+10t+3 equation:

(H)=-2H^2+10H+3

We move all terms to the left:

(H)-(-2H^2+10H+3)=0

We get rid of parentheses

2H^2-10H+H-3=0

We add all the numbers together, and all the variables

2H^2-9H-3=0

a = 2; b = -9; c = -3;
Δ = b2-4ac
Δ = -92-4·2·(-3)
Δ = 105
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{105}}{2*2}=\frac{9-\sqrt{105}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{105}}{2*2}=\frac{9+\sqrt{105}}{4}
$