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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5H^2-4H-10=0

a = 5; b = -4; c = -10;
Δ = b2-4ac
Δ = -42-4·5·(-10)
Δ = 216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6\sqrt{6}}{2*5}=\frac{4-6\sqrt{6}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6\sqrt{6}}{2*5}=\frac{4+6\sqrt{6}}{10}
$