H(5)=t^2-2

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



(5)=H^2-2
We move all terms to the left:
(5)-(H^2-2)=0
We get rid of parentheses
-H^2+2+5=0
We add all the numbers together, and all the variables
-1H^2+7=0
a = -1; b = 0; c = +7;
Δ = b2-4ac
Δ = 02-4·(-1)·7
Δ = 28
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:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{7}}{2*-1}=\frac{0-2\sqrt{7}}{-2} =-\frac{2\sqrt{7}}{-2} =-\frac{\sqrt{7}}{-1} $
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{7}}{2*-1}=\frac{0+2\sqrt{7}}{-2} =\frac{2\sqrt{7}}{-2} =\frac{\sqrt{7}}{-1} $

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