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0.5x^2-2=4
We move all terms to the left:
0.5x^2-2-(4)=0
We add all the numbers together, and all the variables
0.5x^2-6=0
a = 0.5; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·0.5·(-6)
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{3}}{2*0.5}=\frac{0-2\sqrt{3}}{1} =-\frac{2\sqrt{3}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{3}}{2*0.5}=\frac{0+2\sqrt{3}}{1} =\frac{2\sqrt{3}}{1} $
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