H(t)=-0.4t^2+92t+25

Solution for H(t)=-0.4t^2+92t+25 equation:

(H)=-0.4H^2+92H+25

We move all terms to the left:

(H)-(-0.4H^2+92H+25)=0

We get rid of parentheses

0.4H^2-92H+H-25=0

We add all the numbers together, and all the variables

0.4H^2-91H-25=0

a = 0.4; b = -91; c = -25;
Δ = b2-4ac
Δ = -912-4·0.4·(-25)
Δ = 8321
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{-(-91)-\sqrt{8321}}{2*0.4}=\frac{91-\sqrt{8321}}{0.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-91)+\sqrt{8321}}{2*0.4}=\frac{91+\sqrt{8321}}{0.8}
$