H(x)=-0.2x^2+26

Solution for H(x)=-0.2x^2+26 equation:

(H)=-0.2H^2+26

We move all terms to the left:

(H)-(-0.2H^2+26)=0

We get rid of parentheses

0.2H^2+H-26=0

a = 0.2; b = 1; c = -26;
Δ = b2-4ac
Δ = 12-4·0.2·(-26)
Δ = 21.8
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{21.8}}{2*0.2}=\frac{-1-\sqrt{21.8}}{0.4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{21.8}}{2*0.2}=\frac{-1+\sqrt{21.8}}{0.4}
$