H(2b)=8(2b)^2-4

Solution for H(2b)=8(2b)^2-4 equation:

(2H)=8(2H)^2-4

We move all terms to the left:

(2H)-(8(2H)^2-4)=0

We get rid of parentheses

-82H^2+2H+4=0

a = -82; b = 2; c = +4;
Δ = b2-4ac
Δ = 22-4·(-82)·4
Δ = 1316
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:

$\sqrt{\Delta}=\sqrt{1316}=\sqrt{4*329}=\sqrt{4}*\sqrt{329}=2\sqrt{329}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{329}}{2*-82}=\frac{-2-2\sqrt{329}}{-164}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{329}}{2*-82}=\frac{-2+2\sqrt{329}}{-164}
$