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(10)=H^2+6
We move all terms to the left:
(10)-(H^2+6)=0
We get rid of parentheses
-H^2-6+10=0
We add all the numbers together, and all the variables
-1H^2+4=0
a = -1; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-1)·4
Δ = 16
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*-1}=\frac{-4}{-2} =+2 $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*-1}=\frac{4}{-2} =-2 $
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