H(-2)=5x^2-4

Solution for H(-2)=5x^2-4 equation:

(-2)=5H^2-4

We move all terms to the left:

(-2)-(5H^2-4)=0

We add all the numbers together, and all the variables

-(5H^2-4)-2=0

We get rid of parentheses

-5H^2+4-2=0

We add all the numbers together, and all the variables

-5H^2+2=0

a = -5; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-5)·2
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{10}}{2*-5}=\frac{0-2\sqrt{10}}{-10}
=-\frac{2\sqrt{10}}{-10}
=-\frac{\sqrt{10}}{-5}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{10}}{2*-5}=\frac{0+2\sqrt{10}}{-10}
=\frac{2\sqrt{10}}{-10}
=\frac{\sqrt{10}}{-5}
$