H(x)=-2x^2+20x+48

Solution for H(x)=-2x^2+20x+48 equation:

(H)=-2H^2+20H+48

We move all terms to the left:

(H)-(-2H^2+20H+48)=0

We get rid of parentheses

2H^2-20H+H-48=0

We add all the numbers together, and all the variables

2H^2-19H-48=0

a = 2; b = -19; c = -48;
Δ = b2-4ac
Δ = -192-4·2·(-48)
Δ = 745
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{-(-19)-\sqrt{745}}{2*2}=\frac{19-\sqrt{745}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{745}}{2*2}=\frac{19+\sqrt{745}}{4}
$