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0.5x^2-4=20
We move all terms to the left:
0.5x^2-4-(20)=0
We add all the numbers together, and all the variables
0.5x^2-24=0
a = 0.5; b = 0; c = -24;
Δ = b2-4ac
Δ = 02-4·0.5·(-24)
Δ = 48
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*0.5}=\frac{0-4\sqrt{3}}{1} =-\frac{4\sqrt{3}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*0.5}=\frac{0+4\sqrt{3}}{1} =\frac{4\sqrt{3}}{1} $
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