H(x)=-2x^2+2x+40

Solution for H(x)=-2x^2+2x+40 equation:

(H)=-2H^2+2H+40

We move all terms to the left:

(H)-(-2H^2+2H+40)=0

We get rid of parentheses

2H^2-2H+H-40=0

We add all the numbers together, and all the variables

2H^2-1H-40=0

a = 2; b = -1; c = -40;
Δ = b2-4ac
Δ = -12-4·2·(-40)
Δ = 321
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{321}}{2*2}=\frac{1-\sqrt{321}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{321}}{2*2}=\frac{1+\sqrt{321}}{4}
$