H(3)=2x^2-4

Solution for H(3)=2x^2-4 equation:

(3)=2H^2-4

We move all terms to the left:

(3)-(2H^2-4)=0

We get rid of parentheses

-2H^2+4+3=0

We add all the numbers together, and all the variables

-2H^2+7=0

a = -2; b = 0; c = +7;
Δ = b2-4ac
Δ = 02-4·(-2)·7
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{14}}{2*-2}=\frac{0-2\sqrt{14}}{-4}
=-\frac{2\sqrt{14}}{-4}
=-\frac{\sqrt{14}}{-2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{14}}{2*-2}=\frac{0+2\sqrt{14}}{-4}
=\frac{2\sqrt{14}}{-4}
=\frac{\sqrt{14}}{-2}
$