H(t)=10+4t+-5t2

Solution for H(t)=10+4t+-5t2 equation:

(H)=10+4H+-5H^2

We move all terms to the left:

(H)-(10+4H+-5H^2)=0

We use the square of the difference formula

-(10+4H-5H^2)+H=0

We get rid of parentheses

5H^2-4H+H-10=0

We add all the numbers together, and all the variables

5H^2-3H-10=0

a = 5; b = -3; c = -10;
Δ = b2-4ac
Δ = -32-4·5·(-10)
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{-(-3)-\sqrt{209}}{2*5}=\frac{3-\sqrt{209}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{209}}{2*5}=\frac{3+\sqrt{209}}{10}
$