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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2H^2+2H-50=0

a = 2; b = 2; c = -50;
Δ = b2-4ac
Δ = 22-4·2·(-50)
Δ = 404
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:

$\sqrt{\Delta}=\sqrt{404}=\sqrt{4*101}=\sqrt{4}*\sqrt{101}=2\sqrt{101}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{101}}{2*2}=\frac{-2-2\sqrt{101}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{101}}{2*2}=\frac{-2+2\sqrt{101}}{4}
$