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(-3)=3-2H^2
We move all terms to the left:
(-3)-(3-2H^2)=0
We add all the numbers together, and all the variables
-(3-2H^2)-3=0
We get rid of parentheses
2H^2-3-3=0
We add all the numbers together, and all the variables
2H^2-6=0
a = 2; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·2·(-6)
Δ = 48
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}$$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*2}=\frac{0-4\sqrt{3}}{4} =-\frac{4\sqrt{3}}{4} =-\sqrt{3} $$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*2}=\frac{0+4\sqrt{3}}{4} =\frac{4\sqrt{3}}{4} =\sqrt{3} $
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